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- Show that $x^3y^3(x^3+y^3) \leq 2$ 3 answers
Let $x$, $y$ are two positive real numbers such that $x+y=2$. Then show that $x^3y^3(x^3+y^3)≤2$.
I have tried a lot using $AP$ $GP$ inequality formulas but failed. Also proceed like that, $x+y=2$ implies $(x+y)^3=8$ implies $x^3+y^3≤ 8$ since $x$ and $y$ are positive, but failed to prove. Please help me to prove this inequality.