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Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq 2xy(x+y+xy)$ for $x,y \in \mathbb{R}^+$.

I started by multiplying everything out on the RHS to get the equivalent statement \begin{align*} x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq 2 x^2 y + 2 x y^2 + 2 x^2 y^2. \end{align*}

I've tried numerous methods of attacking this problem, but I haven't had much luck yet. The closest I seem to have been able to get is to use the identity $(x-y)^2\geq 0$ to deduce that \begin{align*} x^3+x^3y^2\geq x^3+xy^2=x(x^2+y^2)\geq 2x^2y, \end{align*} and similarly, \begin{align*} y^3x^2+y^3\geq x^2y+y^3=y(x^2+y^2)\geq 2xy^2. \end{align*}

But now I am left with having to prove that $x^2+y^2\geq 2x^2y^2$, which is obviously false for $x,y \in \mathbb{R}^+$. I'm thinking I'm giving up too much trying to make the identity $(x-y)^2\geq 0$ work for this problem.

I also tried writing everything as a polynomial in $x$ and showing that the resulting polynomial is positive for $x>0$, but this gets extremely messy. Does anyone have any ideas? Thanks in advance!

stochasm
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  • Note: the edited condition is not necessary. Your inequality holds for all positive reals, as in the solution you accepted. Please undo your edit. – Calvin Lin May 24 '14 at 15:25

2 Answers2

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By $AM-GM :$

$$x^3+x^3y^2\ge 2x^3y$$

$$x^2+y^2\ge2xy$$ $$y^3 +x^2y^3\ge2xy^3$$

Add : $$\text{LHS}\ge2xy(x^2+y^2+1)$$

WLOG Assume $x\ge y\ge 1$

Use rearrangement inequality on same inequality :

$$x^2+y^2+1\ge x+y+xy$$

By transitivity...

evil999man
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By AM-GM, we have $$x^3+y^3+x^2y^3+x^3y^2 \geq 4 \sqrt[4]{x^3y^3x^2y^3x^3y^2} = 4x^2y^2.$$ Similarly, we have $$x^3y^2 + y^2 + x^2 + x^3 \geq 4 \sqrt[4]{x^3y^2y^2x^2x^3} = 4x^2y$$ and $$x^2y^3+ x^2 + y^2 + y^3 \geq 4 \sqrt[4]{x^2y^3x^2y^2y^3} = 4xy^2.$$ Adding these inequalities yields the desired result.

user133281
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