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# INMO 2017 Practice SET-II (Algebra only)

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

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

Note by Priyanshu Mishra
1 year, 2 months ago

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I suppose that 6 uses the chebyshev's once · 1 year, 2 months ago

This is the proposal so I don't know the answer. · 1 year, 2 months ago

Not started yet. But i ve no idea of the first one pls post the sol or give a hint · 1 year, 2 months ago

I cant understand what u say · 1 year, 2 months ago

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 · 1 year, 2 months ago

Nice, concise solution.

HAVE you solved the problem 6? · 1 year, 2 months ago

Will the other discussions also be present on Brilliant Discussion link as this one · 1 year, 2 months ago

Yes. · 1 year, 2 months ago

Is the ans to Q5 2^1/3 i m doubtful · 1 year, 2 months ago

Yes I think you forgot to write it as 1/cube root 2. · 1 year, 2 months ago

I am in desperate need of geometry and combinatorics sets pls hlp · 1 year, 2 months ago

I will post it before INMO. · 1 year, 2 months ago

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 · 1 year, 2 months ago

The answer to4 is f(x)=0 &f(x)=(-1)^n+1(x(1-x))^n · 1 year, 2 months ago