Prove that $\frac{x^x}{x+y}+\frac{y^y}{y+z}+\frac{z^z}{z+x} \geqslant \frac32$ 
$x,y,z >0$, prove
  $$\frac{x^x}{x+y}+\frac{y^y}{y+z}+\frac{z^z}{z+x} \geqslant \frac32$$

I have a solution for this beautiful and elegant inequality. I am posting this inequality to share and see other solutions from everyone on MSE.

Since some of you asked for my solution, I used the following estimations
$$x^x \geqslant \frac12 \left(x^2+1 \right)$$
Proving rest of this inequality is very simple.
Method 1:
$$\sum_{cyc}\frac{x^x}{x+y} \geqslant \frac12\sum_{cyc}\frac{x^2+1}{x+y}\geqslant \frac12 \left( \frac{(x+y+z)^2}{2(x+y+z)}+\sum_{cyc}\frac{1}{x+y} \right) \geqslant \frac12 \sum_{cyc}\left(\frac{x+y}{4}+\frac{1}{x+y} \right) \geqslant \frac12 \sum_{cyc} 2 \cdot\sqrt{\frac{x+y}{4}\cdot \frac{1}{x+y}} \geqslant \frac32$$
Method 2:
$$\sum_{cyc} \frac{x^x}{x+y} \geqslant \frac12 \sum_{cyc}\frac{x^2+1}{x+y} \geqslant \frac32 \sqrt[3]{\frac{(x^2+1)(y^2+1)(z^2+1)}{(x+y)(y+z)(z+x)}}\geqslant \frac32$$
It is because C-S yields $$(x^2+1)(y^2+1) \geqslant (x+y)^2$$
so $$\frac{(x^2+1)(y^2+1)(z^2+1)}{(x+y)(y+z)(z+x)} \geqslant 1$$  
Method 3: 
Notice that
$$\sum_{cyc} \frac{x^2+1}{x+y}=\sum_{cyc} \frac{y^2+1}{x+y}$$
Hence
$$\sum_{cyc}\frac{x^x}{x+y} \geqslant \frac12 \sum_{cyc}\frac{x^2+1}{x+y}=\frac14\sum_{cyc} \frac{x^2+y^2+2}{x+y} \geqslant\frac14\sum_{cyc}\frac{\frac{(x+y)^2}{2}+2}{x+y}$$$$=\frac14 \sum_{cyc} \left(\frac{x+y}{2}+\frac{2}{x+y} \right) \geqslant \frac14 \sum_{cyc}2 =\frac32$$
Can you find a different solution that does not employ the estimation $x^x \geqslant \frac12(x^2+1)$ ?

Update 1 
The solution of user260822 is not correct. This is not a Nesbitt's inequality. 
We have
$$\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x} > 1 $$
(not $\frac32$)

Update 2
Another similar question I posted a few weeks back 
Unconventional Inequality $ \frac{x^x}{|x-y|}+\frac{y^y}{|y-z|}+\frac{z^z}{|z-x|} > \frac72$
Please help. Thank you
 A: 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 $x=y=1$.
A: 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": $$\sqrt[n]{x_{1}^{x_{1}}x_{2}^{x_{2}}\cdots x_{n}^{x_{n}}}\geq \frac{x_1+x_2+\cdots+x_n}{n}.$$
