RMO algebra DPP 1

 


Let x^k = a.
Then the expression inside sigma is k * a/1 + a^2.
Whenever you see something like a/1 + a^2, think about A.M. >= G.M.
Why?
Since a/1 + a^2 = 1/(a + 1/a).
And a + 1/a >= 2 due to A.M. G.M. when a > 0.
So 
1/(a + 1/a) <= 1/2

So if you unroll the Sigma here, you will get:
1 * x^1/(1 + x^2) + 2 * x^2/(1 + x^4) ....

And for x > 0:
LHS <= 1 * 1/2 + 2 * 1/2 ... n*1/2 = 1/2*[n * (n+1)/2] = n*(n+1)/4

So max value of LHS is what the question is asking for.
It means
A.M. = G.M. and x^k = 1/x^k => x = 1

So only solution for x > 0 is x = 1.

Also, x = 0 is not possible since that will make LHS = 0 whereas RHS > 0.
Now we are left with x < 0 case.

Even when x > 0 each term on the left had to its upper bound to be able to equal the RHS.

Now with x < 0, half the terms are negative and other half positive.
Even when all the positive terms hit their upper bounds, the negative ones will pull them back.
And it will never be able to reach RHS.

So no real solution for x < 0.


This is from RMO 2019 paper.

We will use this identity here:
x^n - y^n = (x - y) (x^(n-1).y^0 + x^(n-2).y^1 ... x^0.y^(n-1))




Let g(x) = x.f(x) - abc
Then g(x) is a cubic with 3 roots a,b,c since f(x) is a quadratic.
=> g(x) = k.(x-a).(x-b).(x-c) = x.f(x) - abc
To find k,
Let x = 0
=> k = 1

So x.f(x) - abc = (x-a)(x-b)(x-c)

Now put put x = a + b + c to find f(a + b + c).




Source





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