inequalites of an acute triangle angles $ 180^{180}*a^b*b^c*c^a \le (a^2+b^2+c^2)^{180} $

Question

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

Answer 1

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.

Answer 2

To solve this you should know (this fact)[https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means]

Hint:

$180 = a + b + c$

Then we can use the fact that: $\frac{a\cdot b + b\cdot c + c\cdot a}{a+b+c} \ge (a+b+c)(a^b\cdot b^c \cdot c^a)^{\frac{1}{a+b+c}} $

After you understood how to use all this facts you will have the inequality:

$\frac{a\cdot b + b\cdot c + a\cdot c}{a+b+c} \le a^2 + b^2 + c^2$

That much more easy to prove.

One more fact that might be helpful:

$\frac{a+b}{2} \le (\frac{a^2 + b^2}{2})^{\frac{1}{2}}$