# Showing that $\left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3$

I would like to show that:

$$\left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3$$

Using AM-GM inequality:

$$\left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3\frac{(a+2b)(b+2c)(c+2a)}{(a+2c)(b+2a)(c+2b)}$$

It suffices to show that:

$$\frac{(a+2b)(b+2c)(c+2a)}{(a+2c)(b+2a)(c+2b)} \geq1$$

$$\Longleftrightarrow ab^2+bc^2+ca^2-(a^2b+b^2c+c^2a)\geq0$$

$$\Longleftrightarrow x+y+z\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

$$xyz=1$$

($x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$)

How can I prove this last inequality? Is there any simpler proof?

-
The last inequality can't possibly be true. If $xyz=1$, then $\frac{1}{x}\frac{1}{y}\frac{1}{z}=1$, and the only way the inequality can hold for both $x,y,z$ and $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ is if $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, which is clearly not true in general. You're going to need to start with something sharper than the AM-GM... –  Micah Aug 29 '12 at 16:19

I will prove that this is true for $a,b,c>0$. If they're allowed to be non-positive, there are counterexamples; e.g., $(a,b,c)=(1,0,-1)$.

Since this expression is homogeneous in $a,b,c$, we may assume without loss of generality that $a+b+c=1$. Then setting $x=c-b$, $y=a-c$, $z=b-a$ yields: $$\left(\frac{a+2b}{a+2c}\right)^3+\left(\frac{b+2c}{b+2a}\right)^3+\left(\frac{c+2a}{c+2b}\right)^3=\left(\frac{1-x}{1+x}\right)^3+\left(\frac{1-y}{1+y}\right)^3+\left(\frac{1-z}{1+z}\right)^3 \, .$$ But if $a,b,c$ are positive and $a+b+c=1$, then $0<a,b,c<1$, from which it follows that $-1<x,y,z<1$. As the function $f(t)=\left(\frac{1-t}{1+t}\right)^3$ is concave upward on $(-1,1)$, we have: $$\left(\frac{1-x}{1+x}\right)^3+\left(\frac{1-y}{1+y}\right)^3+\left(\frac{1-z}{1+z}\right)^3=f(x)+f(y)+f(z) \geq 3f\left(\frac{x+y+z}{3}\right)=3f(0)=3 \, .$$ This completes the proof.

-
If it was concave then the inequality should not be the other way around? –  clark Aug 29 '12 at 18:30
@clark: Concave up (i.e., convex, if you prefer the concave/convex terminology to the concave up/down terminology). –  Micah Aug 29 '12 at 18:37
I am sorry, I didnot know that. (+1) –  clark Aug 29 '12 at 18:58
Thank you for your answer! $a,b,c$ are positive real numbers. I can't see why the condition $a+b+c=1$ can be added. –  Chon Aug 30 '12 at 17:48
@Chon: If $I(a,b,c)$ is the left-hand side of your inequality, then $I(\lambda a, \lambda b, \lambda c)=I(a,b,c)$ for all $\lambda \neq 0$. Take $\lambda=\frac{1}{a+b+c}$... –  Micah Aug 30 '12 at 17:50

Or maybe just do it by Holder's Inequality? $$\left(\sum_{cyc} \left(\frac{a+2b}{a+2c} \right)^3 \right) \left(\sum_{cyc}(a+2b)(a+2c)\right)\left(\sum_{cyc}(a+2c)\right)^2 \ge \left(\sum_{cyc}(a+2b)\right)^4$$ As $\displaystyle \sum_{cyc}(a+2c) = 3(a+b+c) = \sum_{cyc}(a+2b)$, and $$\displaystyle \sum_{cyc}(a+2b)(a+2c) = a^2 + b^2 + c^2 + 8(ab + bc + ca) = (a+b+c)^2 + 6(ab+bc+ca) \le 3(a+b+c)^2$$ because $3(ab + bc+ca) \le (a+b+c)^2$, hence we have that $$\sum_{cyc} \left(\frac{a+2b}{a+2c} \right)^3 \ge 3$$ which finishes the proof.

-
Thank you for your answer Rijul Saini! I don't recognize Holder's inequality in your first line, could you detail the calculations? –  Chon Aug 30 '12 at 17:43
See Holder's Inequality. There, take $\lambda_1 =\lambda_2 =\lambda_3 = 1/5, \lambda_4 = 2/5$, and raise both sides to the 5th power. –  Rijul Saini Aug 30 '12 at 19:04
OK, thank you for the link. –  Chon Sep 1 '12 at 15:11