# Math Help - Inequality

1. ## Inequality

if x+y=2
then show that (x^3+y^3)(x^3)(y^3)<=2

2. Start by breaking $x^{3}+y^{3}$ down.

3. Thanks,got it..

4. You could use the Arithmetic-Geometric mean inequality, for $x,\;y\ \ge\ 0$

$x+y=2$

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

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

$\left(x^3+y^3\right)x^3\;y^3=2(4-3xy)x^3\;y^3\ \le\ 2\;\;?$

$(4-3xy)x^3\;y^3\ \le\ 1\;\;?$

AM-GM Inequality: $\displaystyle\frac{x+y}{2}\ \ge\sqrt{xy}\Rightarrow\ xy\ \le\ 1$

$(4-3xy)x^3\;y^3=\displaystyle\frac{4}{\left(\frac{1}{ x^3\;y^3}\right)}-\frac{3}{\left(\frac{1}{x^2\;y^2}\right)}$

The divisor of 4 is greater than than the divisor of 3.