parmenides51 wrote: Prove that all positive real $x, y, z$ satisfy the inequality $x^y + y^z + z^x > 1$. (D. Bazylev) If one of them is not less than $1$, so it's true. Let $\{x,y,z\}\subset(0,1).$ Thus, $$\sum_{cyc}x^y>\sum_{cyc}\frac{x}{x+y}>\sum_{cyc}\frac{x}{x+y+z}=1.$$

more details pls $$\sum_{cyc}x^y>\sum_{cyc}\frac{x}{x+y}$$

yefangzhou wrote: more details pls $$\sum_{cyc}x^y>\sum_{cyc}\frac{x}{x+y}$$ $$x^y=\frac{1}{\left( \frac{1}{x}\right)^y} =\frac{1}{\left( 1+\frac{1-x}{x}\right)^y} \geqslant \frac{1}{ 1+\left( \frac{1-x}{x}\right) y}=\frac{x}{x+y-xy}>\frac{x}{x+y}.$$

Feridimo wrote: yefangzhou wrote: more details pls $$\sum_{cyc}x^y>\sum_{cyc}\frac{x}{x+y}$$ $$x^y=\frac{1}{\left( \frac{1}{x}\right)^y} =\frac{1}{\left( 1+\frac{1-x}{x}\right)^y} \geqslant \frac{1}{ 1+\left( \frac{1-x}{x}\right) y}=\frac{x}{x+y-xy}>\frac{x}{x+y}.$$ No, the Bernoulli inequality is not correct. Edit: It turns out that i'm wrong.

completely wrong

Illuzion wrote: We have $x^y=(1+(x-1))^y\geq 1+y(x-1),$ It's wrong for $0<y<1$. For $\{x,y\}\subset(0,1)$ we have $$\left(1+\frac{1}{x}-1\right)^y\leq1+y\left(\frac{1}{x}-1\right).$$