# proof of inequality

• Aug 13th 2012, 11:54 PM
albert940410
proof of inequality
$\displaystyle please\,\, have \,\,a \,\,try !$

$\displaystyle (1) a+b+c=1$

$\displaystyle (2)a,b,c \in R^+$

$\displaystyle prove :\,\,27(a^3+b^3+c^3)\geq 6(a^2+b^2+c^2)+1$

$\displaystyle with \,\,greatest \,\, appreciation \,\,! !$
• Aug 14th 2012, 10:09 AM
FernandoRevilla
Re: proof of inequality
Quote:

Originally Posted by albert940410
$\displaystyle (1) a+b+c=1$ $\displaystyle (2)a,b,c \in R^+$ $\displaystyle prove :\,\,27(a^3+b^3+c^3)\geq 6(a^2+b^2+c^2)+1$

One way: consider $\displaystyle f(x,y,z)=27(x^3+y^3+z^3)-6(x^2+y^2+z^2)-1$ defined on the compact set

$\displaystyle K=\{(x,y,z)\in\mathbb{R}^3:x+y+z+1=0,\;x\geq 0,\;y\geq 0,\;z\geq 0\}$

Prove that the absolute minimum of $\displaystyle f$ is $\displaystyle 0$ .
• Aug 14th 2012, 05:50 PM
richard1234
Re: proof of inequality
You could use Lagrange multipliers or a similar technique, but I'm pretty sure this can be solved without calculus (e.g. Muirhead's inequality or AM-GM).