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Let $x,y$ be positive reals such that $x+y=2$. Prove that :

$x^3y^3(x^3+y^3) \leq 2$

Source : INMO 2002

My attempt :

I started with the left side of the inequality to be proved.

$x^3y^3(x^3+y^3) = x^3y^3(x+y)(x^2+y^2-xy) = 2 x^3y^3(x^2+y^2+2xy-3xy)$

$=2x^3y^3(4-3xy)$

How to proceed ?

Do I have to some AM-GM or Cauchy-Schwarz on particular set of values ?

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2 Answers 2

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You are nearly done.

Applying AM-GM gives us that $$ xy \times xy \times xy \times (4-3xy) \le \frac {(4-3xy+xy+xy+xy)^4}{4^4}=1$$

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    O Wow !!! That solved problem.. Thanks +1 – 2017-01-15
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Proceeding afterwards, we need to prove that $$x^3y^3 (4-3xy)\leq 1$$ Now we apply AM-GM inequality to the positive reals $4-3xy, xy, xy, xy $ and we obtain $$(xy)^3 (4-3xy) \leq (\frac {4-3xy+3xy}{4})^4 \leq 1$$ Hope it helps.