If $x,y,z$ are positive real numbers such that $x+y+z=3$, prove that
$$x^xy^yz^z\ \ge\ 1$$
My attempt
This is equivalent to
$$x\ln x + y\ln y + z\ln z\ \ge\ 0$$
By AM-GM
$\dfrac{x\ln x + y\ln y + z\ln z}3 \ge \sqrt[3]{xyz(\ln x)(\ln y)(\ln z)}$
I don't know what to do next.
You can't use AM–GM: some of the terms on the LHS may be negative. Try Jensen's inequality instead.
Hint :
$x \longmapsto x\ln(x)$ is convex. Use convexity inequality aka Jensen's inequality to prove your result.
$1+\ln x$ is injective on $x \in (0,\infty)$ so it follows by Lagrange multipliers, that the minimum of $x^xy^yz^z$ subject to $x+y+z=3$ must occur when $x=y=z$:
From Lagrange multipliers we must solve the system,
$$x^x(1+\ln x)y^yz^z=\lambda$$
$$y^y(1+\ln y)x^xz^z=\lambda$$
$$z^z(1+\ln z)x^xy^y=\lambda$$
This means we must have $1+\ln x=1+\ln y=1+\ln z$.
With the bordered hessian technique you can show that $x=y=z=1$ does indeed give a minimum of $x^xy^yz^z=1$.
Let $f(x)=x\ln{x}$.
Hence, $f''(x)=\frac{1}{x}>0$, which says that $f$ is a convex function.
Thus, by Jensen $$\frac{\sum\limits_{cyc}x\ln{x}}{3}\geq\frac{x+y+z}{3}\ln\frac{x+y+z}{3}=0$$ and we are done!