Find max: $\frac{a}{b+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}$

Question

For $a,b,c>0$ and $abc=1$. Find max: $\frac{a}{b+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}$

Answer 1

Hint1
$2LHS=3-(\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c})$
Hint2
Apply Cauchy-schwarz:

$[ b(b+2a)+c(c+2b)+a(a+2c)][\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c} ]\ge [a+b+c]^2$

Answer 2

We will prove: $\frac{a}{b+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}\le 1$ (1)

$(1)\Leftrightarrow \frac 1{\frac ba+2}+\frac 1{\frac cb+2}+\frac 1{\frac ac+2}\le 1 $. Denoting $x=\frac ba, y=\frac cb, z=\frac ac$, we get

$x,y,z>0; xyz=1; \frac 1{x+2}+\frac 1{y+2}+\frac 1{z+2}\le 1$ (2)

$(2)\Leftrightarrow \frac {\frac 12(x+2)-\frac x2}{x+2}+\frac {\frac 12(y+2)-\frac y2}{y+2}+\frac {\frac 12(z+2)-\frac z2}{z+2}\le 1$

We write the last inequality as $\frac x{x+2}+\frac y{y+2}+\frac z{z+2}\ge 1$ and so it is enough to prove that $\frac x{x+2}\ge \frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}}$ (3)

We have

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

$=x\sqrt[3]{x^2}+2\sqrt[6]{x^6.y^2.z^2}$

$=x\sqrt[3]{x^2}+2\sqrt[3]{x^2}$

$=\sqrt[3]{x^2}(x+2)$

(3) is proved. Of course, we have two other similar inequalities. The equality when $a=b=c=1$ and the maximum value is $1$