inequalites of an acute triangle angles $ 180^{180}*a^b*b^c*c^a \le (a^2+b^2+c^2)^{180} $ If $a,b,c$ are an acute angle of triangle the prove that
$ 180^{180}*a^b*b^c*c^a \le (a^2+b^2+c^2)^{180} $
No idea

Taking $180^{\text{th}}$ roots and using $a+b+c=180^\circ$, the inequality is equivalent to 
$$(a^b b^c c^a)^{\frac1{a+b+c}} \le \frac{a^2+b^2+c^2}{a+b+c}$$
Use weighted AM-GM in the form
$$(a^b b^c c^a)^{\frac1{a+b+c}}  \le \frac{b\cdot a + c\cdot b+a\cdot c}{b+c+a}$$
and $ab+bc+ca \le a^2+b^2+c^2$ to conclude.
