Inequality $a^ab^bc^c \geq (a+b-c)^a(b+c-a)^b(c+a-b)^c$ I found a problem where now I essentially have to show the following inequality
$$a^ab^bc^c \geq (a+b-c)^a(b+c-a)^b(c+a-b)^c$$ where $a,b,c$ are the sides of a triangle.
I have tried a lot of approaches but am not making headway. 
Any hint in the right direction would be highly appreciated.
: Could it be the above inequality is false and what was meant to be proved was what Calvin Lin proved below....If so, can anybody help provide a counterexample?
 A: The inequality to be proven is equivalent to
$$\left(\frac{a+b-c}{a}\right)^{\frac{a}{a+b+c}}\left(\frac{b+c-a}{b}\right)^{\frac{b}{a+b+c}}\left(\frac{c+a-b}{c}\right)^{\frac{c}{a+b+c}}\leq 1\,.$$
By the Weighted AM-GM Inequality,
$$\begin{align}&\left(\frac{a+b-c}{a}\right)^{\frac{a}{a+b+c}}\left(\frac{b+c-a}{b}\right)^{\frac{b}{a+b+c}}\left(\frac{c+a-b}{c}\right)^{\frac{c}{a+b+c}}\\&\phantom{abcde}\leq \frac{a}{a+b+c}\left(\frac{a+b-c}{a}\right)+\frac{b}{a+b+c}\left(\frac{b+c-a}{b}\right)+\frac{c}{a+b+c}\left(\frac{c+a-b}{c}\right)\,.\end{align}$$
Can you guess what the right-hand side of the inequality above is equal to?
A: Hint: Just do the "natural" simplifications.
I'm guessing that you already did the following first step, if you said that you've "tried a lot of approaches".   

 Use the substitution of $ a = y+z-x, b = z+x-y, c = x+y-z$, we want to show that
$$\prod (\frac{2x} { y+z } ) ^ { y+z } \leq 1. $$

Now, the terms seem complicated because of the mixture of terms, which prevents us from applying Jensen's. However, we can remove that difficulty easily.   

 Normalize to $ x + y + z = 1$, and taking log, we want to show that
$$ \sum (1-x) \ln ( \frac{ 2x}{ 1-x })  \leq 0 $$

This follows immediately from Jensen's, with equality when $ x  = y = z = \frac{1}{3} $.  
