Show that $a^{bc}+b^{ca}+c^{ab}>2$ for $a,b,c\in(0,1)$

Question

Show that $a^{bc}+b^{ca}+c^{ab}>2$ for $a,b,c\in(0,1)$.

I have no idea how to attempt this exponential inequality and couldn't find the condition for equality. Any hints will be appreciated.

Answer 1

Let $\{x,y\}\subset(0,1)$.

Hence, by Bernoulli $\left(\frac{1}{x}\right)^y=\left(1+\frac{1}{x}-1\right)^y\leq1+y\left(\frac{1}{x}-1\right)=\frac{x+y-xy}{x}$.

Id est, $\sum\limits_{cyc}a^{bc}\geq\sum\limits_{cyc}\frac{a}{a+bc-abc}$.

Thus, it remains to prove that $\sum\limits_{cyc}\frac{a}{a+bc-abc}\geq2$, which is

$2a^2b^2c^2-2abc+1+\sum\limits_{cyc}(-2a^2b^2c+a^2bc+a^2b+a^2c-a^2)\geq0$ 0r

$$(1-a^2)(1-b^2)(1-c^2)+\sum\limits_{cyc}(a^2b^2c^2-2a^2b^2c+a^2bc)+$$ $$+\frac{1}{2}\sum\limits_{cyc}(a^2b+a^2c-2a^2b^2)+\frac{1}{2}\sum\limits_{cyc}\left(a^2b+a^2c-\frac{4}{3}abc\right)\geq0,$$ which is obvious because $$\sum\limits_{cyc}(a^2b+a^2c-2a^2b^2)=\sum\limits_{cyc}(a^2(b-b^2)+a^2(c-c^2))\geq0,$$ $$\sum\limits_{cyc}(a^2b^2c^2-2a^2b^2c+a^2bc)=abc\sum\limits_{cyc}a(1-b)(1-c)\geq0$$ and $$\sum\limits_{cyc}\left(a^2b+a^2c-\frac{4}{3}abc\right)\geq\sum\limits_{cyc}\left(a^2b+a^2c-2abc\right)=\sum\limits_{cyc}c(a-b)^2\geq0$$ Done!