Prove that $\frac{x^x}{x+y}+\frac{y^y}{y+z}+\frac{z^z}{z+x} \geqslant \frac32$

Solution 1:

Since $e^x\geq 1+x$ for all $x\in \mathbb{R}$, by letting $x=-\log u$ we have that for any $u >0$, $\log u\geq 1-\frac{1}{u}$, and so $u\log u>u-1$ for any $u>0$. Combining these two inequalities, it follows that $$\sqrt{x^x y^y}= \exp\left(\frac{x\log x}{2}+\frac{y\log y}{2}\right)\geq \exp\left(\frac{x+y}{2}-1\right)\geq \frac{x+y}{2}.$$ Hence $$\frac{x^x}{x+y}+\frac{y^y}{y+z}+\frac{z^z}{z+x}\geq \frac{1}{2} \left(\sqrt{\frac{x^x}{y^y}}+\sqrt{\frac{y^y}{z^z}}+\sqrt{\frac{z^z}{x^x}}\right),$$ and the result follows from the AM-GM inequality.

Remark: By combining $e^x\geq 1+x$ for $x\in\mathbb{R}$ and $u\log u\geq u-1$ for $u>0$ as above, we obtain the following "reverse exponential AM-GM inequality": $$\sqrt[n]{x_{1}^{x_{1}}x_{2}^{x_{2}}\cdots x_{n}^{x_{n}}}\geq \frac{x_1+x_2+\cdots+x_n}{n}.$$

Solution 2:

Preliminary note: If $x=1+u$ and $y=1+v$ with some small $u,v$ then $$ \frac{x^x}{x+y} = \frac{(1+u)^{1+u}}{2+u+v} \approx \frac12 \frac{1+u}{1+\frac12u+\frac12v} \approx \frac12\left(1+\frac12u-\frac12v\right) = \frac12 + \frac{x-y}4. $$

We will prove that $$ \frac{x^x}{x+y} \ge \frac12 + \frac{x-y}{4}. \tag1 $$ Then the statement follows immediately.

(1) is equivalent with $$ x^x \ge \frac{x+y}2 + \frac{x^2-y^2}4 = \frac{(x+1)^2}4 - \frac{(y-1)^2}4. $$ The maximum of the RHS is attained at $y=1$, so it suffices to prove $$ x^x \ge \frac{(x+1)^2}4, $$ or equivalently, $$ f(x) = x \ln x - \ln4 -2\ln(x+1)\ge0. $$

We have $f'(x) = 1+\ln x - \frac{2}{x+1}>0$ for $x>1$ and $f'(x)<0$ for $0<x<1$. Therefore, $\min f = f(1)=0$ and $f(x)\ge0$ for all $x>0$. Equality occurs in (1) only for $x=y=1$.