On proving that $(x+y)^z+(x+z)^y+(y+z)^x > 2 $ with $x,y,z > 0$.

Question

I would like to show that $(x+y)^z+(x+z)^y+(y+z)^x > 2 $ with $x,y,z > 0$.

My attempts, I feel, have been going nowhere. How could this be proven?

Answer 1

The hint:

If one of variables is greater or equal to $1$ then it's obvious.

But for $\{x,y,z\}\subset(0,1)$ by Bernoulli we obtain: $$\sum_{cyc}(x+y)^z>\sum_{cyc}\frac{x+y}{x+y+z}=2.$$

I used the following Bernoulli: $$(x+y)^z=\frac{1}{\left(1+\frac{1}{x+y}-1\right)^z}\geq\frac{1}{1+z\left(\frac{1}{x+y}-1\right)}=\frac{x+y}{x+y+z-z(x+y)}>\frac{x+y}{x+y+z}.$$