Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$

Question

$a,b,c \in \mathbb{R}$ 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

$$8^a=(2^a)^3$$so if you name $$2^a=A,2^b=B,2^c=C$$so prove $$A^3+B^3+C^3 >= A+B+C$$and not that A,B,C>0 — Khosrotash, Dec 28 '14 at 09:12

score 3 · Answer 1 · answered Dec 28 '14 at 09:16

3

Hint: $f(x) = x^3 - x$, $g(x) = \dfrac{x^3}{9} - x$ $\text{ increases}$, $x \geq 3$ and Jensen's inequality !

answered Dec 28 '14 at 09:16

DeepSea

77,651

score 1 · Answer 2 · answered Dec 28 '14 at 10:18

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)

score 1 · Accepted Answer · edited May 06 '15 at 07:43

1

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 $$(x+y+z)^2\ge (3\sqrt[3]{xyz})^2=9.$$

edited May 06 '15 at 07:43

Blind

1,116

answered Dec 28 '14 at 10:19

math110

93,304

score 0 · Answer 4 · answered May 06 '15 at 07:33

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}

Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$

4 Answers4