# RMO Practice Board!

Post as many problems as you can for others to try.

Please don't post problems from past year RMO papers.

Have FUN!

Let $a, b, c, d, e, f$ be real numbers such that the polynomial equation

$x^{8} - 4x^{7} +7x^{6} +ax^{5}+ bx^{4} + cx^{3} + dx^{2} + ex + f = 0$

has eight positive real roots. Determine all possible values of $f$ Note by Harsh Shrivastava
5 years ago

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Since we were given that the roots were positive reals,we get motivated and tempted to use,$R.M.S\geq A.M\geq G.M\geq H.M$,also since we have been given,$\sum \alpha$ and $\sum \alpha\beta$ we try to find the value of $\sum \alpha^2=4^2-2\times 7=2$.We apply $R.M.S \geq A.M$ $\sqrt{\dfrac{2}{8}} \geq \dfrac{4}{8}\\ \Longrightarrow \dfrac{1}{2} \geq \dfrac{1}{2}$,hence equality occurs,meaning that all the roots are equal.Now,that means,$A.M=G.M$,hence$\dfrac{4}{8}=\sqrt{f}\\ \Longrightarrow \dfrac{1}{256}=f$,hence there is only one value of $f=\dfrac{1}{256}$.Is this correct Harsh?

- 5 years ago

In your solution you need prove that $(\sum_{cyclic}ω)^{2}-\sum_{cyclic}ω_{1}ω_{2}=\sum_{cyclic}ω^{2}$

- 5 years ago

That's an identity that is true for all numbers, real and imaginary. And there should be a 2 infront of that second summation.

- 4 years, 11 months ago

Yes , its one of the short Newton identities.

- 4 years, 11 months ago

If $x,y,z$ are positive real numbers, prove that,

$\large (x+y+z)^2 (yz+zx+xy)^2 \le 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$

Give me as much different solutions please .

- 5 years ago

U may use the concept of fermat pt to solve this one..then it will be very easy...suppose three line segments AF,BF,CF of length x,y,z meet each other at120 degrees and now use cosine rule to find out the lengths of the triangle and see that each term on the rhs of the inequality is actually squares of the sides...then express xy+yz+zx as $\frac{4}{\sqrt{3}}T$ where T is the area of the triangle....then use $T=\frac{abc}{4R}$ and see that the inequality reduces to 3R greater than or equal to (x+y+z) which is obvious since F is the fermat pt

- 5 years ago

A very elegant solution

- 1 year, 1 month ago

You have a set of consecutive positive integers,$\{1,2,3,...,199,200\}$,if you choose $101$ numbers from these at random,then prove that there at-least two numbers of which one divides the other.

- 5 years ago

If $x$ and $y$ are integers such that $y^2+3(xy)^2=30x^2+517$, Prove that $3(xy)^2=588$

- 5 years ago

Let $y^{2} = b$ and $x^{2} = a$.

$b(1 + 3a) = 30a +517$

$\implies b(1+3a) -10(3a + 1) = 507$

$\implies (b-10)(3a+1) = 507 \times 1 = 13 \times 39 = 3 \times 169$

Since a , b are integers, after checking through all 6 cases, we get only admissible values of $(x ,y) = ( \pm 2, \pm 7)$.

This forces $3(xy)^{2}= 588$.

$\QED$

- 5 years ago

Find all positive integer solutions $a$, $b$ and $c$ such that

$\dfrac {7}{a^2} + \dfrac {11}{b^2} = \dfrac {c}{77}.$

- 5 years ago

$\text{THIS IS ONLY AN OUTLINE OF THE SOLUTION, NOT A COMPLETE SOLUTION.}$

$\dfrac {7 }{a^2} + \dfrac {11 }{b^2} = \dfrac{c}{77}.$

Case 1: $c>1$

This forces $a^{2} < 7 \text{ and } b^{2} < 11$

Now it is easy to check that $(a,b,c) = (1,1,1386) \text{ and } (a,b,c) = (3,3,154)$ satisfy the given conditions.

Case 2: $c \leq 77$

Since $c \leq 77$ this implies $\dfrac{c}{77} \leq 1$.

Let $\dfrac{c}{77} = \dfrac{1}{k}$ for some integer $k \geq 1$.

$\implies c = \dfrac{77}{k}$

Now since $c$ is a positive integer , this forces $k \text{ | } 77$.

Possible values for $k = 1,7,11,77$.

Now plugging the possible values of $c$ in the given equation and using Simon's Favourite Factoring Trick, we can easily solve for $a,b$

- 5 years ago

Find all primes $p$ such that for all prime $q$, the remainder of $p$ upon divison by $q$ is squarefree.

- 5 years ago

I think that 3 is the only prime number. I used the fact that all primes>3 is of the form $\pm 1 \pmod{6}$.

x, y, z are positive real numbers such that x^2+y^2+z^2=3 Prove that x+y+z >= (xy)^2+(yz)^2+(zx)^2

Sorry for not using latex as I was in a hurry.

- 5 years ago

Can u send the solution of this question ? I have been trying. No progress ....

- 5 years ago

I didn't get it.

- 5 years ago

Let $x,y$ and $z$ be positive real numbers such that $x+y+z=1$. Prove that $\frac{yz}{1+x}+\frac{zx}{1+y}+\frac{xy}{1+z} \le \frac{1}{4}$.

I got this. This is simple AM-HM inequality. Write 1+x=(1-y)+(1-z). Similarly for others

Apply AM-HM and then manipulate on the RHS.

- 5 years ago

Find all reals $x, y, z$ in (1,∞) such that

$x + y + z + \dfrac{3}{x-1}+ \dfrac{3}{y-1} + \dfrac{3}{z-1} = 2(\sqrt{x+2} + \sqrt{y+ 2} + \sqrt{z+2})$

- 5 years ago

- 5 years ago

If $a,b,c$ are positive integers such that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}<1$. Prove that the maximum value of $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$ is $\dfrac{41}{42}$

If none of $a,b,c=2$, the maximum value of the expression will be $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}<\frac{41}{42}$. So, one of them has to $2$ say $a$. So we have to prove that the maximum value of $\frac{1}{b}+\frac{1}{c}$ is $\frac{10}{21}$. If none of $b,c=3$, the maximum of the expression will be $\frac{1}{4}+\frac{1}{5}=\frac{9}{20}<\frac{10}{21}$. So, one them has to be $3$ say $b$. Therefore, $\frac{1}{c} \le \frac{1}{7}$.Therefore the maximum value of $c$ is $7$ and the maximum value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is $\frac{41}{42}$.

Not sure if this is the best method.

Find all primes $x,y$ such that $x^{y}-y^{x}=xy^{2}-19$.Enjoy solving this.

- 5 years ago

Using FLT, we have

$x^y \equiv x \pmod{y} \Rightarrow x \equiv -19 \pmod {y} \Rightarrow x+19 \equiv 0 \pmod{y}$

$y^x \equiv y \pmod{x} \Rightarrow -y \equiv -19 \pmod {x} \Rightarrow 19-y \equiv 0 \pmod{x}$

From here, it is clear that either $x=2$ or $y=2$. We now separate this into 2 cases:

Case 1: $y=2$

When $y=2$, we have $19 - 2 = 17 \equiv 0 \pmod{x}$, which implies $x=17$. The solution in this case is $(17, 2)$. Checking, however, we find that this clearly doesn't work. So, there are no solutions in this case.

Case 2: $x=2$

When $x=2$, we have $19+2 = 21 \equiv 0 \pmod {y}$, which implies $y=3$ or $y=7$. Checking, we find both these solutions work. Thus, 2 solutions exist in this case: $(2, 3)$ and $(2,7)$.

Therefore, only 2 solutions exist: $(2, 3)$ and $(2,7)$.

- 5 years ago

would you please explain how x = -19 mody

- 5 years ago

Putting $x^y-y^x=xy^2-19$ in $\pmod{y}$, we get $x=-19 \pmod{y}$.

- 5 years ago

Using FLT and modulo $x$ and $y$ we get that,$x+19 \equiv 0 \pmod{y},y-19 \equiv 0 \pmod{x}$,now just use parity(as suggested by @Sharky Kesa).

- 5 years ago

Let $p$ and $q$ be distinct primes. Show that

$\displaystyle \sum_{j=1}^{(p-1)/2}[\frac{qj}{p}]+ \sum_{j=1}^{(q-1)/2}[\frac{pj}{q}]=\dfrac{(p-1)(q-1)}{4}$.

Note that $[ \ ]$ represents the greatest integer function.

Let $S={(x,y)|x,y \in N, \quad 1 \le x \le (p-1)/2, \quad 1 \le y \le (q-1)/2}$.

We note that the set $S$ can be partitioned into two sets $S_1$ and $S_2$ according as $qx>py$ or $qx. Note that there are no pairs in such $S$ such that $qx=py$.

The set $S_1$ can be described as the set of all pairs $(x,y)$ satisfying $1 \le x \le (p-1)/2, \quad 1 \le y < qx/p$. Then $S_1$ has $\displaystyle \sum_{j=1}^{(p-1)/2}{\frac{qj}{p}}$ elements.

Similarly, $S_2$ has $\displaystyle \sum_{i=1}^{(q-1)/2}{\frac{pi}{q}}$. Hence, we get the required result.

Let $a,b,c,d \in \mathbb{N}$ such that $a \ge b \ge c \ge d$ and $d \neq 0$. Show that the equation $x^4 - ax^3 - bx^2 - cx -d = 0$ has no integer solution in $x$.

Please post a "complete solution" like you would do in RMO.

- 5 years ago

I'll have a crack at it. As $d \neq 0$, $x \neq 0$.

Denote $y = \frac{1}{x}$.

We have

$x^4 = ax^3 + bx^2 + cx + d \quad \quad$

$1 = ay + by^2 + cy^3 + dy^4 \quad \quad (1)$

Now, we separate this into 2 cases:

Case 1:

$x = -n$ where $n \in \mathbb{N}$, then $y=-\dfrac{1}{n}$. Substituting this into $(1)$, we get

$\dfrac {-a}{n} + \dfrac {b}{n^2} + \dfrac {-c}{n^3} + \dfrac {d}{n^4} = 1$

$\dfrac {-an+b}{n^2}+\dfrac{-cn+d}{n^4}=1$

Here, the LHS is non-positive since $a \geq b$ and $c \geq d$, whereas the RHS is positive. Thus no solutions exist in this case.

Case 2:

$x = n$ where $n \in \mathbb{N}$, then $y=\dfrac{1}{n}$. Substituting this into $(1)$, we get

$\dfrac {a}{n} + \dfrac {b}{n^2} + \dfrac {c}{n^3} + \dfrac {d}{n^4} = 1$

This implies that $n \neq 1$, and

$\dfrac {a}{n} < 1$

$1 \leq \dfrac {a}{n} + \dfrac {a}{n^2} + \dfrac {a}{n^3} + \dfrac {a}{n^4} < \displaystyle \sum_{i=1}^{\infty} \dfrac{a}{n^i} = \dfrac {a}{n-1}$

These two equations imply that

$\dfrac {n}{a} > 1$

$\dfrac {n-1}{a} < 1$

There are no such $n$ that satisfy the above two inequalities. Thus, no solutions exist.

- 5 years ago

Isn't this from INMO 2013 ??

- 4 years, 11 months ago

Yes it is.

- 4 years, 11 months ago

Hey why aren't u participating in the INMO practice board ?

- 4 years, 11 months ago

let n be an integral solution (if possible). then n [n^3-an^2-bn-c] = d. Now for the equation to hold n divides d. so n is less than or equal to d. If n is negative -ax^3 is positive and is greater than bx^2 as a is greater than b.similarly for the next term and the total value becomes positive but known the value is 0.hence our assumption that n is negative is wrong .so n is positive and less than d. Now as n<d <a, n^4<an^3 and -bn^2 and -cx and -d all are negative,so the sum is also negative which was again 0. hence contradiction. So no integral roots exists.

- 7 months, 3 weeks ago

Let $a,b,c$ be positive reals satisfying $(a+b)(b+c)(c+a) = 1$.

Prove that $ab+bc+ca \leq \frac{3}{4}$

- 5 years ago

Using am gm

(a + b) + (b + c) + (c + a) > 3

a + b + c > 3/2

now using cauchy,

$3(a^2 + b^2 + c^2) > (a + b + c)^2$

so

$a^2 + b^2 + c^2 > 3/4$

using cauchy again,

$(a^2 + b^2 + c^2)^2 > (ab + bc + ca)^2$

3/4 > ab + bc + ca

- 5 years ago

Nice solution. You could also have used the fact that $(a+b+c)^{2} \ge 3(ab+bc+ca)$

Dev,your solution is incorrect. Check the solution again,you will find your flaw.

- 5 years ago

You got the following two inequalities,$a^2+b^2+c^2>ab+bc+ca\\ and\\ a^2+b^2+c^2>\dfrac{3}{3}$,what next?How can you conclude that,$\dfrac{3}{4}>ab+bc+ca$?

- 5 years ago

using cauchy again, like this

(sum(a^2))(sum(b^2)) > (sum(ab))^2

sum is cyclic

- 5 years ago

If $a,b \ge 0$. Prove that $\sqrt{2}(\sqrt{a(a+b)^{3}}+b \sqrt{a^{2}+b^{2}}) \le 3(a^{2}+b^{2})$.

Prove that $2 < (\dfrac{n+2}{n+1}) ^ {(n+1)}$ for all integers $n > 0$.

- 5 years ago

Well its easy just use Bernoulli's inequality that (1+x)^n > 1+nx

- 5 years ago

Let $p>3$ be a prime number. Suppose $\displaystyle \sum_{k=1}^{p-1}{\frac{1}{k}}=\frac{a}{b}$ where $\gcd(a,b)=1$. Prove that $a$ is divisible by $p^{2}$.

I am able to prove that a is divisible by p. But not able to prove divisible by p^2. Could u send the solution ?

- 5 years ago

- 4 years, 12 months ago

My proof is incomplete I could not prove that a is divisible by p^2

- 4 years, 12 months ago

Post the proof that a is divisible by p, Thanks!

- 4 years, 12 months ago

Take LCM on the LHS and try proving that each term in the numerator leaves a different residue modulo p. Do it by contradiction. Once you are able to prove that, the rest is clear. If it is difficult, first take a smaller example say p=7. Then work it out

- 4 years, 12 months ago

I have the solution only till $a$ is divisible by $p$. To prove that $a$ is divisible by $p^{2}$ is given below the solution as a challenge.

In a triangle $ABC$ , D is the midpoint of side AB and E is the point of trisection of side BC nearer to C.

Given that $\angle ADC = \angle BAE$, find $\angle BAC$.

- 5 years ago

Is the answer angle BAC = 90 degrees ? I hope so My proof : Apply Menelaus theorem on triangle BCD with transversal EA. Obtain that F is the midpoint of CD. Then its simple angle chasing :)

- 5 years ago

Shrihari,we get that $CF=2DF$

- 5 years ago

I don't think so check again.

- 5 years ago

Applying Menelaus' theorem, ( (BE/EC)(CF/FD)(DA/AB)=1

- 5 years ago

BE/EC is 2 and DA/AB is 1/2. So CF/FD is 1

- 5 years ago

Yup I also got 90 degrees.

- 5 years ago

Yes , Menelaus is indeed a good approach for this problem. For shortening the solution: Once you get F is the midpoint of CD , you can directly say that $\angle BAC = 90^\circ$ , do you see why?

- 5 years ago

FD=FC=FA . Great !

- 5 years ago

Do u have some problems for practice ? I need some number theory. I am not that good or in other words I am bad in it

- 5 years ago

I also need some NT probs...

- 5 years ago

Even I need NT problems...

- 5 years ago

Are you going to appear for RMO from mumbai region or from the pune region ?

- 5 years ago

Pune region.

- 5 years ago

There are given $2^{500}$ points on a circle labeled $1, 2, . . . , 2^{500}$in some order. Prove that one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums of the pairs of numbers at the endpoints of the chosen chords are equal.

- 5 years ago

Given seven arbitrary distinct real numbers, show that there exist two numbers $x$ and $y$ such thst $0<\dfrac{x-y}{1+xy}<\dfrac{1}{\sqrt{3}}$.

The question is easy. Consider x and y to be some tan(A) and tan(B) respectively and consider A and B to vary from 0 to pi so that all reals are covered. now divide it into six equal portions spaced by pi/6. Now u have 6 pigeons and 7 holes and you are done. Then that expression is simply tan(A-B).

- 5 years ago

If a circle goes through point $A$ of a parallelogram $ABCD$, cuts the two sides $AB,AD$ and the diagonal $AC$ at points $P,R$ and $Q$ respectively. Prove that $(AP×AB)+(AR×AD)=AQ×AC$.

I hope you all have drawn the diagram.(Consider all the segments below as vectors in the given direction.)

Let $AS$ be the diameter of given circle.

$AB+AD=AC$

Therefore taking dot product with$AS$

$AB.AS+AD.AS=AC.AS$

Using, $a.b=|a|*|b|*cos(\theta)$ [$\theta$ is angle between the vectors]

$|AB|*|AP|+|AR|*|AD|=|AQ|*|AC|$

- 5 years ago

Let $ABC$ be a triangle and $P$ be a point inside it.Let $x,y,z$ denote the lengths of $PA$,$PB$,$PC$.Let $a,b,c$ be the lengths of sides $BC$,$CA$,$AB$.Find all points $P$ such that $axy+byz+czx=abc$.

- 5 years ago

Find the least positive integer n such that 2549 divides $n^{2545} - 2541$

- 5 years ago

Please post the solution , Thanks!

- 5 years ago

Let $ABC$ be an acute-angled tiangle .Let O denote its circumcenter.Let $\Pi$ be the circle passing through the points A,O,B.Lines CA and CB meet $\Pi$ again at P and Q respectively.Prove that PQ is perpendicular to the line CO.

- 5 years ago

- 5 years ago

Hint: Let $\overrightarrow{QP} \cap \overleftrightarrow{CO} = \{R\}$ and $\overleftrightarrow{CO} \cap \Pi = \{S\}$ and then it will suffice to prove that $\Delta CRP \sim \Delta CAS$

- 5 years ago

An easy INMO geometry problem:

Let $ABC$ be a triangle with circumcircle $\Gamma$. Let $M$ be a point in the interior of triangle $ABC$ which is also on the bisector of $\angle A$. Let $AM, BM, CM$ meet $\Gamma$ in $A_{1}, B_{1}, C_{1}$ respectively. Suppose $P$ is the point of intersection of $A_{1}C_{1}$ with $AB$ and $Q$ is the point of intersection of $A_{1}B_{1}$ with $AC$. Prove that $PQ$ is parallel to $BC$ .

- 5 years ago

Let $a,b \in R, a \neq 0$. Show that $a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{b}{a} \ge \sqrt{3}$.

Well the question becomes very easy if u make it a quadratic in b and then prove that the Discriminant is less than 0. I did it that way.

- 5 years ago

If $a,b,c,d,e$ are real numbers, prove that the roots of $x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e=0$ cannot all be real if $2a^{2}<5b$.

Let roots be $p,q,r,s,t$.

Sum of roots = -a & Product of roots taken two at a time = b.

Therefore Sum of squares of roots = $a^{2} - 2b$

If all roots are real, then by Titu's Lemma,

$a^{2} -2b \geq \frac{a^{2}}{5}$

$\implies 2a^{2} \geq 5b$

Therefore the polynomial cannot have all roots real if $5b > 2a^{2}$.

- 5 years ago

Show that $n^{5}+n^{4}+1$ is not prime for $n>1$.

$n^{2} + n + 1$ is factor of the given expression.Hence the result is obvious.

- 5 years ago

Prove that

$1<\frac { 1 }{ 1001 } +\frac { 1 }{ 1002 } +.........+\frac { 1 }{ 3001 } <\frac { 4 }{ 3 }$

- 5 years ago

It is easy. I got a calculus and a non calculus answer

- 5 years ago

Since Rmo is a precalculus olympiad, answer without calculus will be better:P

- 5 years ago

I think we can use calculus in rmo .But if we use it we need to be very careful in our steps and not miss any points.

- 5 years ago

Can u tell me how to add an image in the answer?

- 5 years ago

First use an image uploading website (I prefer postimg.org ), upload your picture, and copy the image link. The format for uploading images in a solution is Alternate text. Type the text which you wish to be displayed if the image doesn't load in the third brackets, and paste the image URL in the second brackets. In the following example, the image URL is http://s12.postimg.org/5qhlgca71/untitled.png, and I wrote the code This is an image file :D. Here's how it appears:

this is a copy paste from this note on moderators

- 5 years ago

If $m$ is a positive real number and satisfies the equation $2m=1+\frac{1}{m}+\sqrt{m^{2}-\frac{1}{4}}$ Find $m$.

- 5 years ago

Could you send the solution of this question. I tried expanding and landed up with a fourth degree polynomial whose roots I am unable to guess

- 5 years ago

Find all pairs $(m,n)$ such that $2^m+3^n$ is a perfect square. (A popular NT problem)

- 5 years ago

Its easy.Only one solution (2,2)

- 4 years, 12 months ago

Wrong. Check again.

- 4 years, 12 months ago

I think he means (4,2).

- 4 years, 12 months ago

Ya I meant (4,2). Sorry,typing error. :P

- 4 years, 12 months ago

- 4 years, 12 months ago

Taking mod 4 we get that m,n should be even. Then using $gcd({2}^{m},3^{n})=1$, we use the general formula for pythagorean triplets. Taking various cases,we get that only $m=4,n=2$ are the solutions.

- 4 years, 12 months ago

Let d(k) denote the sum of the digits of k in base10. Find all natural numbers n such that n, d(n), d(d(n)) sum to 2015.

- 5 years ago

I am getting no solution. Is that right ? My proof : n,d(n),d(d(n)) are congruent to the same number say 'x' (mod 9). So their sum is congruent to 3x (mod 9). Therefore 2015 is congruent to 3x ( mod 9) which is impossible as 3 does not divide 2015

- 5 years ago

Here is a geometry question. Waiting for numerous solutions. In a triangle ABC, excircle opposite to A touches BC at point Q. Let M be the midpoint of BC. Incircle of triangle ABC touches BC at point P. Prove that PM=MQ. I have been trying this question but I am stuck. So please solve this question.

- 5 years ago

Its easy , note that $BP=QC=s-b$ where $s$ is the semiperimeter of the triangle and $b$ is length of side opposite to $\angle B$ respectively. Since $M$ is mid point of $BC$ , the result follows.

- 5 years ago

Are protractors allowed for RMO? Nothing about it is mentioned in the website and the hall ticket.

In the instructions given in question paper , it is said that protractors are not allowed.

- 5 years ago

Oh nice !

- 5 years ago

Try this set guys.

- 5 years ago

$1991^{k}|(1990^{1991^{1992}}+1992^{1991^{1990}})$

$max(k)=?$

Details:

$b|a\rightarrow$ $a$ is divisible by $b$.

- 5 years ago

1991

- 5 years ago

How ? Could u give some rough outline of your solution ?

- 5 years ago

Shrihari,I have posted the general result of that combinatorics problem in its 'discuss solution' part. Now try to prove the general result .

https://brilliant.org/problems/triangular-grid-problem/?group=U3Rx70QSpE0z&ref_id=1044906

- 5 years ago

Hey Saarthak I haven't heard of Catalan numbers before. You need to elaborate a bit Could you give a proof for the number of ways ? It will help me understand better. Thanks in advance

- 4 years, 12 months ago

Yes..... Can you post solution ?

- 5 years ago

Everyone here who have posted questions on this board and have not got any solutions please post the solutions now as RMO is only 42 hours from now. Its better that we understand the methods. Sorry that I cannot send the solution of my geometry question as I myself have been trying that question but could not arrive at the solution.

And please post some more questions if possible :)

- 5 years ago

Done :)

Thanks

- 5 years ago

Let $x,y$ be real numbers such that $x,y > 1$

Prove that $\dfrac{x^{2}}{y-1} + \dfrac{y^{2}}{ x-1} \geq 8$

- 4 years, 12 months ago

Just use AM-GM inequality directly on the expression in the LHS and then use the fact that (a-2)^2>0 Then rearrange a bit. You get the required answer

- 4 years, 12 months ago

Alternatively we can substitute p = x-1 and q = y-1.

Then it will be easy manipulate the expression.

- 4 years, 12 months ago

Nice idea!

- 4 years, 12 months ago

Since the expression is symmetric, let's assume without loss of generality that $x \ge y$.

$\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge \frac{y^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 4+\frac{y^{2}}{x-1} \implies \frac{x^{2}}{y-1} \ge 4$. Minimum value of $\frac{x^{2}}{y-1}$ is $4$ and it occurs if and only if $x=y=2$.

Therefore, $\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 8$.

- 4 years, 12 months ago

Hmm you have not proven that min value of $\frac{x^{2}}{y-1}$ is 4.

- 4 years, 12 months ago

Do you mean that I haven't proven that minimum value of $\frac{y^{2}}{y-1}$ is $4$?

- 4 years, 12 months ago

Sorry didn't see this post but since I found it, here you go - 65 bugs are placed at different squares of a 9*9 square board. Abug in each moves to a horizontal or vertical adjacent square. No bug makes 2 horizontal or vertical moves in succession. Show that after some moves, there will be at least 2 bugs in the same square.

- 5 years ago

$a,b,c$ are real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=1$. Prove that $a^{2}b^{2}c^{2}≤\frac{1}{54}$.

- 5 years ago

See my note

- 5 years ago

A barrel contains $2n$ balls, numbered $1$ to $2n$. Choose three balls at random, one after the other, and with the balls replaced after each draw.

What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, is repeated?

1/2

- 5 years ago

Write the sum $S_n= \displaystyle \sum_{k=0}^n{\frac{(-1)^{k}{n \choose{k}}}{k^{3}+ 9k^{2}+26k+24}}$ in the form $\frac{p(n)}{q(n)}$ where $p(n),q(n)$ are polynomials with integer coefficients.
We first note that $k^{3}+9k^{2}+26k+24=(k+2)(k+3)(k+4)$