Hello friends, try these problems and post solutions :

\((1)\) For each positive integer \(n\), show that there exists a positive integer \(k\) such that

\(k = f(x){ (x + 1) }^{ 2n }+ g(x)({ x }^{ 2n }+ 1)\)

for some polynomials \(f\), \(g\) with integer coefficients, and find the smallest such \(k\) as a function of \(n\).

(\(2)\) Prove that there exists a polynomial \(P(x, y)\) with real coefficients such that \(P(x, y) \ge 0\) for all real numbers \(x\) and \(y\), which cannot be written as the sum of squares of polynomials with real coefficients.

\((3)\) Let \({ x }_{ 1 } = 2\) and \({ x }_{ n + 1 } = { x }^{ 2 }n - { x }_{ n } + 1\) for \(n\ge 1\).

Prove that

\(\large\ 1 - \frac { 1 }{ { { 2 }^{ 2 } }^{ n - 1 } } < \frac { 1 }{ { x }_{ 1 } } + \frac { 1 }{ { x }_{ 2 } } +...+ \frac { 1 }{ { x }_{ n } } < 1 - \frac { 1 }{ { { 2 }^{ 2 } }^{ n } }\).

\((4)\) Find, with proof, all nonzero polynomials \(f(z)\) such that \(f({ z }^{ 2 }) + f(z)f(z + 1) = 0\).

\((5)\) Find the least real number \(r\) such that for each triangle with side lengths \(a\), \(b\), \(c\),

\(\large\ \frac { max(a, b, c) }{ \sqrt [ 3 ]{ { a }^{ 3 } + { b }^{ 3 } + { c }^{ 3 } + 3abc } } < r\).

\((6)\) Let \(a\), \(b\), \(c\) be positive real numbers. Prove that

\(\large\ { \left( \sum { \frac { a }{ b+c } } \right) }^{ 2016 } \le \left( \sum { \frac { { a }^{ 2016 } }{ { b }^{ 2016 } + { b }^{ 2015 }c } } \right) { \left( \sum { \frac { a }{ c + a } } \right) }^{ 2015 }\).

## Comments

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TopNewestI suppose that 6 uses the chebyshev's once – Himanshu Singh · 1 year ago

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– Priyanshu Mishra · 1 year ago

This is the proposal so I don't know the answer.Log in to reply

Not started yet. But i ve no idea of the first one pls post the sol or give a hint – Himanshu Singh · 1 year ago

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– Himanshu Singh · 1 year ago

I cant understand what u sayLog in to reply

I just got the 3 one.its as follows prove by induction that the given expressions equals1-( (x1x2.......xn)^-1) then make the reqd changes in the inequality. By the way the above induction also showds that xn=(x1x2.......xn-1)+1 Now again induction on the transformed ineq completes the sol – Himanshu Singh · 1 year ago

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HAVE you solved the problem 6? – Priyanshu Mishra · 1 year ago

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Will the other discussions also be present on Brilliant Discussion link as this one – Himanshu Singh · 1 year ago

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– Priyanshu Mishra · 1 year ago

Yes.Log in to reply

Is the ans to Q5 2^1/3 i m doubtful – Himanshu Singh · 1 year ago

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– Priyanshu Mishra · 1 year ago

Yes I think you forgot to write it as 1/cube root 2.Log in to reply

I am in desperate need of geometry and combinatorics sets pls hlp – Himanshu Singh · 1 year ago

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– Priyanshu Mishra · 1 year ago

I will post it before INMO.Log in to reply

In the he eq put x0 such that f(x0)=0giving that f(x0^2)also equals to zerozero thus by induction there are infinitely many roots of funless x0=0/1 infinitely many roots gives f(x)=0The other case gives after some calc that f(x)=0 for 0,1 only thus the other function is of the formx^m(x-1)^nkeeping in mind that the leading coefficient must be negative keep the above poly in the given funct eq to obtain n=m some thought to the leading coefficient gives the -1at the beginning – Himanshu Singh · 1 year ago

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The answer to4 is f(x)=0 &f(x)=(-1)^n+1(x(1-x))^n – Himanshu Singh · 1 year ago

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– Priyanshu Mishra · 1 year ago

Correct. Can you show the solution?Log in to reply