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

Let $x=2^a,y=2^b,z=2^c$. Then $x,y,z>0$ and
$$ xyz=2^{a+b+c}=1.$$
By Holder's inequality
$$(x^3+y^3+z^3)(1+1+1)(1+1+1)\ge (x+y+z)^3.$$
Therefore,
$$x^3+y^3+z^3\ge\dfrac{(x+y+z)^3}{9}\ge (x+y+z)$$
because use AM-GM inequality
$$(x+y+z)^2\ge (3\sqrt[3]{xyz})^2=9.$$
