<s>[INST] 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?
 [/INST] 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?
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