<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] We have
$$
8^a+1+1\geq3\sqrt[3]{8^a}=3\times 2^a,
$$
$$
8^b+1+1\geq3\sqrt[3]{8^b}=3\times 2^b,
$$
$$
8^c+1+1\geq3\sqrt[3]{8^c}=3\times 2^c,
$$
$$
2^a+2^b+2^c\geq 3\sqrt[3]{2^{a+b+c}}=3.
$$
It follows that
\begin{eqnarray}
8^a+8^b+8^c&\geq&3(2^a+2^b+2^c)-6\\
&=&(2^a+2^b+2^c)+2(2^a+2^b+2^c-3)\\
&\geq&2^a+2^b+2^c.
\end{eqnarray}
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