proof of inequality

**albert940410** · August 13th 2012, 11:54 PM

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

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

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

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

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

**FernandoRevilla** · August 14th 2012, 10:09 AM

Originally Posted by albert940410

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

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

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

Prove that the absolute minimum of $f$ is $0$ .

**richard1234** · August 14th 2012, 05:50 PM

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).

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