An interesting inequality

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An interesting inequality

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karis

#1 h Oct 16, 2010, 3:58 PM • 2 Y

Y by Adventure10, Mango247

Let $x, y: 0<y\leq x\leq 1$ .Prove that:
$\frac{x^3y^2+y^3+x^2}{x^2+y^2+1}\geq xy$

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kuing

#2 h Oct 16, 2010, 4:25 PM • 2 Y

Y by Adventure10, Mango247

karis wrote:

Let $x, y: 0<y\leq x\leq 1$ .Prove that:
$\frac{x^3y^2+y^3+x^2}{x^2+y^2+1}\geq xy$

because $0<y\leq x\leq 1$ , we can let $x=\frac{a}{a+b},y=\frac{a}{a+b+c}$ , where $a>0$ and $b,c\ge 0$ .

then $x^3y^2+y^3+x^2-xy(x^2+y^2+1)$ $=\frac{a^4 \left(b^2+b c+c^2\right)+a^3 \left(b^3+3 b^2 c+4 b c^2+c^3\right)+a^2b c (b+c)^2}{(a+b)^3 (a+b+c)^3}\ge0$

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