<s>[INST] Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$ $a,b,c \in \mathbb{R}$ and $a+b+c=0$.
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$
I think that $2^{a}.2^{b}.2^{c}=1$, but i don't know what to do next
 [/INST] As it was said, you can take $x=2^a$, $y=2^b$ and $z=2^c$ to obtain:
$$
x^3+y^3+z^3\ge x+y+z
$$
With $x,y,z>0$ and $xyz=1$. It is equivalent to:
$$
x^2(x-1)+y^2(y-1)+z^2(z-1)\ge 0 \iff \frac{x^2(x-1)+y^2(y-1)+z^2(z-1)}{3}\ge 0
$$
Since $x^2$ and $x-1$ ar equally ordered, we might apply Chebychevs inequality to obtain:
$$
\frac{x^2(x-1)+y^2(y-1)+z^2(z-1)}{3}\ge \frac{x^2+y^2+z^2}{3}\frac{(x-1)+(y-1)+(z-1)}{3}
$$
And to show that the two factors are greater than zero isn't difficult. (AM-GM)
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