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 $\sqrt { 3 }$.

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

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## Comments

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TopNewest@Siddharth G :- many congrats for clearing RMO!!!. How many problems did you solve?

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

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Read This. I should say, I didn't do as well as you did. (Regardless of the paper's standard)

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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?

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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.

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Thanks. Now i understood that.

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.

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Are 'Problem primer for the olympiads' and 'Number theory-problems, structures and examples'(by titu .A.) not fine for RMO/INMO ?

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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 $(x-3)(x-1)^2$.

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$2x^2-2x+3$. Though I havent found a good solution yet

The answer seems to beLog in to reply

You can tell me your approach, irrespective of good or bad.

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$2x+1$ when divided by $(x-1)^2$ ,

I took a cubic which left a remainder$2x^3-4x^2+4x+1 = (x-1)^2(x+2) + 2x+1$. 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 $(x-1)^2(x-3)$.

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$f(1)=3$ and $f(3)=15$.

Got a better solution from my friend- Clearly,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)

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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?

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$q_1(x)$ as a whole is a polynomial and in this case the quotient.(ie $q_1 of x$ akin to $f(x)$

That is not multiplication.Log in to reply

Also give me a clue to solve this problem:

Find all real numbers $x$ for which

${ 10 }^{ x }\quad +\quad { 11 }^{ x }\quad +\quad { 12 }^{ x }\quad =\quad { 13 }^{ x }\quad +\quad { 14 }^{ x }$

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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:

$\large\ \frac { a }{ { b }^{ 2 }{ c }^{ 2 } } +\frac { b }{ { c }^{ 2 }{ a }^{ 2 } } +\frac { c }{ { a }^{ 2 }{ b }^{ 2 } } \ge \quad \frac { 9 }{ a+b+c }$

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

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

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

$\large\ \quad =\frac { 9 }{ a+b+c }$

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As you have cleared RMO, can you tell me which book you preferred specially for Geometry and functional equations?

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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 .

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Please help me to factorise this one:

$\large\ { x }^{ 2 }+4{ y }^{ 2 }-2xy-2x-4y-8=0$

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Hello Siddharth G,

Appeared in RMO 2015? How was it? What is your expected score?

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4 Questions. How was yours?

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Actually I am in K.V JNU so i cannot appear in RMO directly. I have to qualify JMO, which i gave this year but unfortunately did not make it. I will try next year( class 11th).

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?

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What is the your probability of clearing RMO?

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As you have qualified RMO 2014, can you tell me that is the INMO camps conducted in delhi or no camp is conducted?

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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

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Have you solved this RMO question in the exam? :

Show that there are infinitely many triples $(x, y, z)$ of integers such that $x^3 + y^4 = z^{31}$.

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$x=-a^4, y=a^3$ for some natural a.

Put z=0 andLog in to reply

Before getting this, what steps you did?

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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.

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$3^n<3^{n-1} \cdot 5$

The numerator is clearly greater than the denominator. For the second part of the inequality note thatLog in to reply

Why it chose 1 and 5 only?

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Are the RMO results out?

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Were they geometry ones?

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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?

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

Please elaborate the steps.

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Hope that you qualify INMO 2016 also, with flying colours.

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$f(x)$ 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 $f(x)$.

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

LetLog in to reply

$f '(1)$ and $f ''(1)$?

What is meant byIntegration or differentiation?

Thanks, but I am unable to understand your solution.

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$f '(1)$ and $f ''(1)$ as $f(1) = a + b + c + d + e + f$?

How can i differentiateAccording 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?

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$a$ of $f(x)$ has multiplicity $m$ if $(x-a)^m | f(x)$ and $(x-a)^{m+1}$ does not divide f(x).

A rootLog in to reply

Sorry for your inconvenience.

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Can you tell me how to start?

Also thanks a lot for sending INMO 2015 notes.

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

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$a+b+c+d+e+f=1 \\ -a+b-c+d-e+f=-1 \\ 5a+4b+3c+2d+e=0 \\ 5a-4b+3c-2d+e=0 \\ 20a+12b+6c+2d=0 \\-20a+12b-6c+2d=0$

Indeed. I made a mistake in my calculations. This is the system you got right?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?

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The arithmetic mean of a pair wise distinct prime numbers is 27. Determine the biggest prime among them.

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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 $a^2.b^2.c^2$.

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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.

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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.

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Should I need to login on the geogebra?

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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.

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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.

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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.

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Have you solved question 1 of INMO?

Also help me in these questions:

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Also you share this problem with your friends of school and FiitJee.

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Also is it possible for you to send me the notes of INMOTC conducted in Delhi and Hyderabad?

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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.

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By the way, which camp was good - DELHI or HYDERABAD?

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This is the list of JMO 2015 qualifiers:

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By the way, in which class are you?

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Find all positive integers $(a, b)$ such that $\large\ \frac { \sqrt { 2 } + \sqrt { a } }{ \sqrt { 3 } + \sqrt { b } }$ is a rational number.

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$(3,2)$ 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 $(3,2)$

For integers,Log in to reply

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When will you send the assignments to my mail?

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