in an given inequality prove that all sides are equal

Question

If $x,y,z$ are the sides of a triangle and $$\left(1+\frac{y-z}{x}\right)^{x}\left(1+\frac{z-x}{y}\right)^{y}\left(1+\frac{x-y}{z}\right)^{z}\geq 1,$$

then prove that $x=y=z$

what i try

assume $\displaystyle f(x)=x\ln\bigg(\frac{k}{x}\bigg)$ and $k>0$

and $\displaystyle f'(x)=-(1+\ln x)$ and $\displaystyle f''(x)=-\frac{1}{x}<0$ for $x>0$

how do i solve it please see it

Answer 1

By AM-GM: $$\sum_{cyc}\frac{x}{x+y+z}\left(1+\frac{y-z}{x}\right)\geq\prod_{cyc}\left(1+\frac{y-z}{x}\right)^{\frac{x}{x+y+z}},$$ which gives $$1\geq\prod_{cyc}\left(1+\frac{y-z}{x}\right)^{\frac{x}{x+y+z}}$$ or $$1\geq\prod_{cyc}\left(1+\frac{y-z}{x}\right)^x$$ and we got a reversed inequality.