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
 A: Hint: $f(x) = x^3 - x$, $g(x) = \dfrac{x^3}{9} - x$ $\text{ increases}$, $x \geq 3$ and Jensen's inequality !
A: 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)
A: 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
$$(x+y+z)^2\ge (3\sqrt[3]{xyz})^2=9.$$
A: 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}
