$\displaystyle please\,\, have \,\,a \,\,try !$

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

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

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

$\displaystyle with \,\,greatest \,\, appreciation \,\,! !$

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- Aug 13th 2012, 11:54 PMalbert940410proof 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 AMFernandoRevillaRe: proof of inequality
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 PMrichard1234Re: 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).