I was doing the the following problem-

**Prove that**\( \frac { \sqrt { a+b+c } +\sqrt { a } }{ b+c } +\frac { \sqrt { a+b+c } +\sqrt { b } }{ c+a } +\frac { \sqrt { a+b+c } +\sqrt { c } }{ a+b } \ge \frac { 9+3\sqrt { 3 } }{ 2\sqrt { a+b+c } } \).

I normalized this to \(a+b+c=1\) and simplified to get-

\(\frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9+3\sqrt { 3 } }{ 2 }\)

By Titu's lemma,

\(\Rightarrow \quad \frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9 }{ 3-(\sqrt { a } +\sqrt { b } +\sqrt { c } ) } \)

However the \(RHS\) is maximized when \(\sqrt { a } +\sqrt { b } +\sqrt { c } \) is maximized which is at \(\sqrt { 3 } \).

\(\Rightarrow \quad \frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9 }{ 3-(\sqrt { a } +\sqrt { b } +\sqrt { c } ) } \le \frac { 9 }{ 3-\sqrt { 3 } } =\frac { 9+3\sqrt { 3 } }{ 2 } \)

What went wrong?

You can view the rest of the problems here

## Comments

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TopNewestHello Siddharth G,

Appeared in RMO 2015? How was it? What is your expected score? – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

4 Questions. How was yours?Log in to reply

By the way, how was your preparation this year?

Did you study something more this year than what you studied last year for RMO 2014? – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

Nothing special, just deeper into the usual topics.Log in to reply

– Priyanshu Mishra · 1 year, 1 month ago

Was the paper harder than last year?Log in to reply

– Siddharth G · 1 year, 1 month ago

A bit, but not harder than average.Log in to reply

What is the your probability of clearing RMO? – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

I prefer not to think about it. Waiting for the results.Log in to reply

As you have qualified RMO 2014, can you tell me that is the INMO camps conducted in delhi or no camp is conducted? – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

Sorry, didnt see the comment. An INMO camp is conducted int IIT Delhi for RMO awardeesLog in to reply

Which topics were taught to you?

If possible, can you send me some notes of that camp to my e-mail?

E-mail- priyanshu_2feb@rediffmail.com – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

I can send the assignments, but my notes are terrible.Log in to reply

– Kulesh Vandan · 4 months ago

Can you please send me your notes and assignments on - kuleshvandan8@gmail.comLog in to reply

Have you solved this RMO question in the exam? :

Show that there are infinitely many triples \((x, y, z)\) of integers such that \(x^3 + y^4 = z^{31}\). – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

Put z=0 and \(x=-a^4, y=a^3\) for some natural a.Log in to reply

Before getting this, what steps you did? – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

The exponent 31 seemed awkward and misplaced. Thus it seemed that it had no effective value.Log in to reply

– Priyanshu Mishra · 1 year, 1 month ago

Can't we use the FMID- variant in this question?Log in to reply

– Siddharth G · 1 year, 1 month ago

FMID is usually used for proving that no solutions exist. However, ther are infinitely many non-trivial solutions as well.Log in to reply

I was doing this question:

Find all positive integers \(n\) such that \(3^{n - 1} + 5^{n - 1}\) divides \(3^n + 5^n\).

The solution in the book was:

Note that \(\large\ 1 < \frac { { 3 }^{ n } + { 5 }^{ n } }{ { 3 }^{ n - 1 } + { 5 }^{ n - 1 } } <5\), so we can have only \(\large\ \frac { { 3 }^{ n } + { 5 }^{ n } }{ { 3 }^{ n - 1 } + { 5 }^{ n - 1 } } \in \{ 2,3,4\}\)

cases, which are easily checked.

Can you explain me why the solution wrote \(\large\ 1 < \frac { { 3 }^{ n } + { 5 }^{ n } }{ { 3 }^{ n - 1 } + { 5 }^{ n - 1 } } <5\) above ?

I am not understanding how that expression came. – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

The numerator is clearly greater than the denominator. For the second part of the inequality note that \(3^n<3^{n-1} \cdot 5 \)Log in to reply

Why it chose 1 and 5 only? – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

1 is just a lower bound by noticing that the numerator is greater than the denominator. For 5, it might help to see that a n tends to infinity, the expression tends to 5Log in to reply

Are the RMO results out? – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

noLog in to reply

Were they geometry ones? – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

Did 1,3,4,6.Log in to reply

Show that there are infinitely many positive real numbers \(a\) which are not integers such that \(a(a-3\text{{a}})\) is an integer.

Which method or theorem you applied? – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

My method is too long. You should check out some other solutions. Eg A family of infinite solutions is odd+0.5.Log in to reply

I am unable to draw it even after trying 5 times?

Please elaborate the steps. – Priyanshu Mishra · 1 year, 1 month ago

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– Siddharth G · 1 year, 1 month ago

http://postimg.org/image/cafnb0sc5/Log in to reply

how you drew the image of the rmo question on the website which you sent me?

Please tell the website where I can draw the diagrams. – Priyanshu Mishra · 1 year ago

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– Siddharth G · 1 year ago

http://web.geogebra.org/. Though it is preferable that you you should at least try to draw it on your own first. PS Sorry for the late response. Couldnt solve your previous problem. Do you have a solution?Log in to reply

No I don't have solution to that inequality .

Please tell me how to save the diagrams of geogebra in desktop as soon as possible. – Priyanshu Mishra · 1 year ago

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– Siddharth G · 1 year ago

Under menu-- export--pngLog in to reply

Should I need to login on the geogebra? – Priyanshu Mishra · 1 year ago

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– Siddharth G · 1 year ago

I dont think so. On the top right corner, (its like an equivalence sign with horizontal bars), go to export and select png. this works on the web version. An alternate method is to take a screenshot with web apps like Nimbus.Log in to reply

By the way how's your preparation for INMO? Was INMOTC conducted at IIT?

If possible, please send me the assignments. You can take as many days you can. – Priyanshu Mishra · 1 year ago

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– Siddharth G · 1 year ago

No INMOTC this year.Log in to reply

IIT professors don't take it initiative. They are just regional coordinators for name.

In KV, 20 DAYS camp is conducted but rarely a student is selected in INMO, where students from delhi region attend no camp, yet so many qualify. GREAT IRONY!!

Also thanks for helping me to post geometric figures. The method worked.

How was your INMO? Can you send me the paper?

Also, clear my doubts of 2 questions i have posted - "Help residues mod 7 continents".

Please post solutions also. – Priyanshu Mishra · 1 year ago

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– Raushan Sharma · 11 months, 3 weeks ago

This time a greater number of students will qualify INMO from the KVS region also. I believe 2 or 3.Log in to reply

– Siddharth G · 11 months, 3 weeks ago

How do you know that?Log in to reply

Find all positive integers \((a, b)\) such that \(\large\ \frac { \sqrt { 2 } + \sqrt { a } }{ \sqrt { 3 } + \sqrt { b } }\) is a rational number. – Priyanshu Mishra · 11 months, 3 weeks ago

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– Raushan Sharma · 11 months, 3 weeks ago

However, if it is asked to find solutions not only in integers but also in rational numbers, then we can generalize a solution to get infinitely many solutionsLog in to reply

– Raushan Sharma · 11 months, 3 weeks ago

For integers, \((3,2)\) is the only solution. Just rationalize the numerator and the denominator, and then note that the radical part of each term of the numerator should be 6. So, fix \(a = 2l^2\) and \(b = 2k^2\) and then simplify to get \((1+l)(1-k) = 0\). And hence you get the only solution as \((3,2)\)Log in to reply

– Raushan Sharma · 11 months, 3 weeks ago

I qualified JMO and KVS RMO this year and appeared for INMO via KVS region, so I met them at INMOTC and also after the exam, I asked some, who I know can qualify.Log in to reply

Also is it possible for you to send me the notes of INMOTC conducted in Delhi and Hyderabad? – Priyanshu Mishra · 11 months, 3 weeks ago

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– Raushan Sharma · 11 months, 3 weeks ago

As per the notes are concerned, I am not that good at taking down notes. However, I can send you the two KVS RMO papers within this week if you give me your e-mail ID. I will send after 2 or 3 days coz we are having ASL tomorrow and the day after. BTW, which class are you in? And did you appear for JMO this year?Log in to reply

I am presently in class 10. Yes, i appeared in JMO in 2015 but could not qualify as i scored just \(48\) marks out of 100.

If you cannot send the notes, then can you atleast send me the assignments provided during the camp?

You can take as many days you want. – Priyanshu Mishra · 11 months, 3 weeks ago

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– Raushan Sharma · 11 months, 3 weeks ago

How did you get to know your marks?? The excel sheet with everyone's marks is not yet out, only those who qualified have got to know their marks.Log in to reply

By the way, which camp was good - DELHI or HYDERABAD? – Priyanshu Mishra · 11 months, 3 weeks ago

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– Raushan Sharma · 11 months, 3 weeks ago

Hyderabad one was better I thinkLog in to reply

– Priyanshu Mishra · 11 months, 3 weeks ago

I doubt. If you qualified JMO, then your name must be there in the pdf format list and the marks are given there.Log in to reply

– Raushan Sharma · 11 months, 3 weeks ago

There is a list of selected students for INMO in KVS's website now, my name is there at 16th serial no. :3Log in to reply

This is the list of JMO 2015 qualifiers:

– Priyanshu Mishra · 11 months, 3 weeks agoLog in to reply

– Raushan Sharma · 11 months, 3 weeks ago

What is there at 31 no.?? See clearly :3Log in to reply

– Priyanshu Mishra · 11 months, 3 weeks ago

Yes that's your name, so why are you asking your marks of JMO as it is already given there?Log in to reply

– Raushan Sharma · 11 months, 3 weeks ago

No, I was not asking mine. I was just telling as ur name is not there, how did you come to know your marks?? One friend of mine also rang up Lawania sir, but he didn't tell the marks!!Log in to reply

By the way, in which class are you? – Priyanshu Mishra · 11 months, 3 weeks ago

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– Raushan Sharma · 11 months, 3 weeks ago

In 10thLog in to reply

– Raushan Sharma · 11 months, 3 weeks ago

It's there, see clearlyLog in to reply

– Priyanshu Mishra · 11 months, 2 weeks ago

When are you sending the KVS RMO papers?Log in to reply

– Siddharth G · 1 year ago

http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYS85LzlkNDNhMWMzNGM3ZDdmNWQ4ZGE3ZDI2NjIxYzg2MThkNjQ2MWM1LmpwZw==&rn=U2Nhbl8yMDE2MDExNy5qcGc= It was okay. Asfor the camp, last time we had one professional olympiad teacher (he gave the notes), one prof from IIT and the rest were past INMO awardees, who studied in IITD. This time the problem(I think) was mainly due to lack of time.Log in to reply

– Priyanshu Mishra · 1 year ago

Suppose i have opened NIMBUS on one side and on the webpage i have opened the geogebra, where i have drawn the image. Then how can i take the snapshot of the image by nimbus?Log in to reply

– Siddharth G · 1 year ago

I solved ~3 questions. In nimbus select "selected area".Log in to reply

– Priyanshu Mishra · 1 year ago

WHERE IS ITS LOCATION IN NIMBUS?Log in to reply

– Siddharth G · 1 year ago

You are using nimbus screenshot (in chrome) right?Log in to reply

– Priyanshu Mishra · 1 year ago

YES.Log in to reply

– Siddharth G · 1 year ago

https://whatsonmypc.files.wordpress.com/2014/10/nimbus.jpeg?w=393&h=494Log in to reply

Can you tell me what is meant by the test "SMT" conducted in all DPSs? I got to know about it from my fiitjee friend who studies at DPS RKP. – Priyanshu Mishra · 12 months ago

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– Siddharth G · 12 months ago

DPS conducts an annual exam for class nine students in partnership with Sof. It is in PCMB.Log in to reply

Have you solved question 1 of INMO?

Also help me in these questions:

– Priyanshu Mishra · 12 months agoLog in to reply

– Siddharth G · 12 months ago

Did Q1 with coordinate geometry. Already tried, couldnt do it.Log in to reply

Also you share this problem with your friends of school and FiitJee. – Priyanshu Mishra · 12 months ago

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– Priyanshu Mishra · 1 year ago

OK. How many questions you solved in INMO?Log in to reply

– Priyanshu Mishra · 1 year, 1 month ago

Thanks for the image. Can you please now elaborate the solution here.?Log in to reply

– Siddharth G · 1 year, 1 month ago

http://www.isical.ac.in/~rmo/rmo(underscore)2015(underscore)wb(underscore)questions(underscore)and(underscore)solutions(underscore)u(underscore)1.pdf. Replace (underscore) by _Log in to reply

– Priyanshu Mishra · 1 year, 1 month ago

You can do problems for INMO 2016 from the set " INMO 2016 PRACTICE SET-1" posted by me . It consists only of 6 number theory problems.Log in to reply

– Siddharth G · 1 year, 1 month ago

thanks!Log in to reply

Hope that you qualify INMO 2016 also, with flying colours. – Priyanshu Mishra · 1 year ago

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– Siddharth G · 1 year ago

Thanks! I hope so too, but it seems unlikely.Log in to reply

\(f(x)\) is a fifth degree polynomial. It is given that \(f(x) + 1\) is divisible by \((x - 1)^3\) and \(f(x) - 1\) is divisible by \((x + 1)^3\). Find \(f(x)\). – Priyanshu Mishra · 1 year ago

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– Raushan Sharma · 11 months, 3 weeks ago

Hey, I remember this. I have solved this one in some NMTC set on brilliant just using simple differential.Log in to reply

– Siddharth G · 1 year ago

Let \( f(x)=ax^5+bx^4+cx^3+dx^2+ex+f \). Note that \( f(1)-1=f(-1)+1=f'(1)=f'(-1)=f''(1)=f''(-1)=0\). You have six variables and six equations.Log in to reply

Integration or differentiation?

Thanks, but I am unable to understand your solution. – Priyanshu Mishra · 1 year ago

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– Siddharth G · 1 year ago

differentiationLog in to reply

According to my knowledge we can differentiate a function only w.r.t some variable , but here are 6 variables.

Could you elaborate the differentiation here? – Priyanshu Mishra · 1 year ago

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– Siddharth G · 1 year ago

differentiate f(x) and then substitute x=1 into f '(x).Log in to reply

– Priyanshu Mishra · 1 year ago

Understood but what is differentiation of a constant? 1 or 0?Log in to reply

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– Priyanshu Mishra · 1 year ago

Which point number?Log in to reply

– Priyanshu Mishra · 1 year ago

I saw that but no such problem is there. On second page of polynomials there are points numbered 4, 5, 6 but i am unable to find. Can you tell the point number?Log in to reply

– Siddharth G · 1 year ago

not a problem. See point 4Log in to reply

– Priyanshu Mishra · 1 year ago

What is meant by multiplicity?Log in to reply

– Siddharth G · 1 year ago

A root \(a\) of \(f(x)\) has multiplicity \(m\) if \((x-a)^m | f(x) \) and \((x-a)^{m+1}\) does not divide f(x).Log in to reply

Sorry for your inconvenience. – Priyanshu Mishra · 1 year ago

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– Siddharth G · 1 year ago

The whole concept might be a bit difficult without a grasp of differentiation. But you can still solve the problem, if you know how to differentiate a polynomial.Log in to reply

Can you tell me how to start?

Also thanks a lot for sending INMO 2015 notes. – Priyanshu Mishra · 1 year ago

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– Siddharth G · 1 year ago

a=9/4, c=-5/4 rest all are zeroLog in to reply

– Priyanshu Mishra · 1 year ago

But \(a=-3/8)\ \(c=5/4\) (e=-15/8)\ and all rest 0 is the correct answer.Log in to reply

– Siddharth G · 1 year ago

Indeed. I made a mistake in my calculations. This is the system you got right? \(a+b+c+d+e+f=1 \\ -a+b-c+d-e+f=-1 \\ 5a+4b+3c+2d+e=0 \\ 5a-4b+3c-2d+e=0 \\ 20a+12b+6c+2d=0 \\-20a+12b-6c+2d=0 \)Log in to reply

I have found another solution without calculus(working half an hour on it). It is completely algebraic.

Here it is:

Let \(f(x) = (x - 1)^3(a{x}^2 + bx + c) - 1\). \(...(1)\)

As \((x + 1)^3|f(x) - 1\) we have

\(f(x) - 1 = a{x}^5 + x^4(b - 3a) + x^3(c - 3b + 3a) + x^2(3b - 3c - a) + x(3c - b) - c - 2\).

and the RHS is divisible by \((x + 1)^3\) , so by dividing RHS by \((x + 1)^3\)

We have the remainder \((-38a - 6c +18b)x^2 + (-48a + 16b)x - 2c - 18a +6b - 2 = 0\)

So, we have three simultaneous equations viz:

\(38a + 6c -18b = 0\) ; \(48a - 16b = 0\) ; \(2c + 18a - 6b + 2 = 0\)

which on solving together gives \(a = \frac {-3}{8}\) ; \(b = \frac {-9}{8}\) ; \(c = -1\).

Putting these values in \((1)\) and doing some tedious calculations we get

\(f(x) = \boxed{\frac { -3 }{ 8 } { x }^{ 5 }+ \frac { 5 }{ 4 } { x }^{ 3 } - \frac { 15 }{ 8 } x}\) which is the correct answer.

Can you please rate my solution? – Priyanshu Mishra · 1 year ago

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– Priyanshu Mishra · 1 year ago

I received assignments of Functional equations and geometry. Are there any assignments on NUMBER THEORY?Log in to reply

– Siddharth G · 1 year ago

No. Though I have a collection of Number Theory Questions from Mumbai RMO.Log in to reply

– Priyanshu Mishra · 1 year ago

Ok. Have you received any e-mail regarding INMOTC 2016?Log in to reply

– Siddharth G · 1 year ago

No.Log in to reply

The arithmetic mean of a pair wise distinct prime numbers is 27. Determine the biggest prime among them. – Priyanshu Mishra · 1 year ago

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– Siddharth G · 1 year ago

The question is terribly framed. I am not sure but I think it means the AM of two different primes is 27. Fins thw maximal possible value of one of the primes.Log in to reply

Which inequality will you apply to solve this ?:

a, b, c are real numbers such that their sum is \(0\) an sum of their square is 1. Find the maximum value of \(a^2.b^2.c^2\). – Priyanshu Mishra · 1 year ago

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– Priyanshu Mishra · 1 year ago

It seems unlikely (seems likely by me) , but with some little efforts it is also possible.Log in to reply

– Priyanshu Mishra · 1 year ago

You can also post solutions there if you have any.Log in to reply

– Priyanshu Mishra · 1 year, 1 month ago

Sorry, i found that.Log in to reply

When will you send the assignments to my mail? – Priyanshu Mishra · 1 year, 1 month ago

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Nice solution!

The second last line should be; "RHS is maximised when √a+ √b+ √c is 'minimised' " as the whole fraction is maximised when its denominator is minimised.

However your deduction is very perfect.

In which class presently you are? – Priyanshu Mishra · 1 year, 4 months ago

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– Siddharth G · 1 year, 4 months ago

11th As for the second last line, \(\sqrt{a} +\sqrt{b} +\sqrt{c}\) when maximized, yields \(3-(\sqrt{a} +\sqrt{b} +\sqrt{c})\) to be minimized,which implies that the RHS is maximized.Log in to reply

I am also preparing for RMO. So as an elder please guide me that which book will be helpful for me to clear RMO?

Also i get to know that you have cleared RMO , so please tell me which book helped you a lot. – Priyanshu Mishra · 1 year, 4 months ago

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– Siddharth G · 1 year, 4 months ago

A lot of it was luck (including the level of the paper). Regardless I found Challenge and Thrill of Pre-College Mathematics and 'Mathematical Circles' to be good at the basic level. Once you have (sufficiently) completed them go for Problem Solving Strategies(Arthur Engel) and books by Titu Andreescu. And keep trying Olympiad questions all the time.Log in to reply

Are 'Problem primer for the olympiads' and 'Number theory-problems, structures and examples'(by titu .A.) not fine for RMO/INMO ? – Priyanshu Mishra · 1 year, 4 months ago

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– Siddharth G · 1 year, 4 months ago

Titu is a good book. Even though Problem Primer is recommended by a lot of people, I did not like it that much.Log in to reply

Please help me in solving this question ( KVS JMO 2015)

A polynomial \(f(x)\) with rational coefficients leaves remainder \(15\), when divided by \(x-3\) and remainder \(2x+1\), when divided by \((x-1)^2\). Find the remainder when \(f(x)\) is divided by \((x-3)(x-1)^2\). – Priyanshu Mishra · 1 year, 4 months ago

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– Siddharth G · 1 year, 4 months ago

The answer seems to be \(2x^2-2x+3\). Though I havent found a good solution yetLog in to reply

You can tell me your approach, irrespective of good or bad. – Priyanshu Mishra · 1 year, 4 months ago

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\(2x^3-4x^2+4x+1 = (x-1)^2(x+2) + 2x+1\). It left a remainder

31when divided by \(x-3\). Note that \((x-1)^2\) leaves a remainder of4when divided by \(x-3\).Thus, \(2x^3-4x^2+4x+1 - 4(x-1)^2 =2x^3-8x^2+12x-3\) leaves a remainder of \(31-4(4)=15\). Hence we find the remainder of \(2x^3-8x^2+12x-3\) when divided by \((x-1)^2(x-3)\). – Siddharth G · 1 year, 4 months ago

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Then, \(f(x)=q_1(x)(x-3)(x-1)^2+r_1(x)\), where \(r_1(x)=ax^2+bx+c\).

Dividing by \((x-1)^2\) and equating remainders,

\(2x+1=0+(ax^2+bx+c-a(x-1)^2)\) [Multiplied by

ato remove the coeff. of \(x^2\)]\(\Rightarrow 2x+1=(2a-b)x+(c-a)\) \(\Rightarrow 2=2a-b, 1=c-a\)

Solving with \(f(1)=3=a+b+c -(1)\)

(a=2, b=-2, c=3) – Siddharth G · 1 year, 4 months ago

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I have your 2nd solution.

What is the problem that i am not able to understand that why you have multiplied x with q1 ( 2nd line) in the polynomial?

You have taken the polynomial as

\(f(x)\quad =\quad { q }_{ 1 }(x)(x-3){ (x-1) }^{ 2 }\quad +\quad { r }_{ 1 }(x)\)

I am asking that why you have multiplied x in above polynomial as it is not given? – Priyanshu Mishra · 1 year, 4 months ago

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– Siddharth G · 1 year, 4 months ago

That is not multiplication. \(q_1(x)\) as a whole is a polynomial and in this case the quotient.(ie \(q_1 of x\) akin to \(f(x)\)Log in to reply

Also give me a clue to solve this problem:

Find all real numbers \(x\) for which

\({ 10 }^{ x }\quad +\quad { 11 }^{ x }\quad +\quad { 12 }^{ x }\quad =\quad { 13 }^{ x }\quad +\quad { 14 }^{ x }\) – Priyanshu Mishra · 1 year, 3 months ago

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– Siddharth G · 1 year, 3 months ago

Try dividing by a suitable number on both sides of the equation.Log in to reply

Please tell me that can i apply CAUCHY - SCHWARZ lemma to prove this inequality ? -

Given \(a, b, c\) are positive real numbers such that

\({ a }^{ 2 }\quad +\quad { b }^{ 2 }\quad +\quad { c }^{ 2 }\quad =\quad 3abc\).

Prove that:

\(\large\ \frac { a }{ { b }^{ 2 }{ c }^{ 2 } } +\frac { b }{ { c }^{ 2 }{ a }^{ 2 } } +\frac { c }{ { a }^{ 2 }{ b }^{ 2 } } \ge \quad \frac { 9 }{ a+b+c }\) – Priyanshu Mishra · 1 year, 3 months ago

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– Siddharth G · 1 year, 3 months ago

Make the numerators on the LHS to the fourth power and then apply the lemma.Log in to reply

As you have cleared RMO, can you tell me which book you preferred specially for Geometry and functional equations? – Priyanshu Mishra · 1 year, 3 months ago

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– Siddharth G · 1 year, 3 months ago

I am terrible at geometry, so I would not recommend anything for that. As for functional equations, the key is to do a lot of problems after learning the basic techniques. I have not found a 'perfect' book, but the good ones are by BJ Venkatachala, Christopher G Small and Problem solving strategies by Arthur Engel.Log in to reply

Any professor or you did by yourself or anyone else?

Also without geometry how can one dream for winning RMO or further olympiads? It is the whole and sole . – Priyanshu Mishra · 1 year, 3 months ago

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– Siddharth G · 1 year, 3 months ago

Did it by myself. I have to make-do without geometry. I am doing geometry (in Challenge and thrill) but it seems futile.Log in to reply

Please help me to factorise this one:

\(\large\ { x }^{ 2 }+4{ y }^{ 2 }-2xy-2x-4y-8=0\) – Priyanshu Mishra · 1 year, 3 months ago

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– Yash Kumar · 1 year, 1 month ago

well this can be simplified by completing the squares.Log in to reply

– Priyanshu Mishra · 1 year, 1 month ago

OK but please explain it here.Log in to reply

– Raushan Sharma · 11 months, 3 weeks ago

Multiply both sides by 2 and try completing the square method.Log in to reply

\(\large\ \frac { { a }^{ 4 } }{ { a }^{ 3 }{ b }^{ 2 }{ c }^{ 2 } } +\frac { { b }^{ 4 } }{ { b }^{ 3 }{ c }^{ 2 }{ a }^{ 2 } } +\frac { { c }^{ 4 } }{ { c }^{ 3 }{ a }^{ 2 }{ b }^{ 2 } } \ge \frac { 9 }{ a+b+c }\)

\(\large\ \frac { { a }^{ 4 } }{ { a }^{ 3 }{ b }^{ 2 }{ c }^{ 2 } } +\frac { { b }^{ 4 } }{ { b }^{ 3 }{ c }^{ 2 }{ a }^{ 2 } } +\frac { { c }^{ 4 } }{ { c }^{ 3 }{ a }^{ 2 }{ b }^{ 2 } } \ge \frac { { ({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }(a+b+c) }\)

\(\large\ \quad \ge \frac { { 9({ abc }) }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }(a+b+c) }\)

\(\large\ \quad =\frac { 9 }{ a+b+c }\) – Priyanshu Mishra · 1 year, 3 months ago

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@Siddharth G :- many congrats for clearing RMO!!!. How many problems did you solve? – Krishna Ar · 2 years ago

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– Siddharth G · 2 years ago

Thanks! I attempted 5 questions, expected to get 4-4.5. How was your paper? (GMO right?)Log in to reply

Read This. I should say, I didn't do as well as you did. (Regardless of the paper's standard) – Krishna Ar · 2 years ago

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