Prove $\left(x^y + y^x \right)\left( \frac{1}{x} + \frac{1}{y} \right)\ge 4$ For arbitrary $x, y > 0$, prove
$$
\left(x^y + y^x \right) \left( \frac{1}{x} + \frac{1}{y} \right) \ge 4
$$
By plotting, it seems true and somewhat tight, but I cannot find a proof for it.

If $x\ge 1,\ y \ge 1$ or $x\le 1,\ y\le 1$, we have
$$
\left(x^y + y^x \right) \left( \frac{1}{x} + \frac{1}{y} \right)\ge \left(x^1 + y^1 \right) \left( \frac{1}{x} + \frac{1}{y} \right) \ge 4
$$
So it suffices to show the inequality for the case $0 < x < 1,\ y > 1$.
Fixing $y$, I tried
$$f(x) = \left(x^y + y^x\right) (x + y) - 4xy$$
and find that $f''(x) > 0$, so we can obtain a minimal point $x_0$ on $(0,1)$ such that $f'(x_0) = 0$. Then we need to prove $f(x_0)\ge 0$. But I have no idea for further progress.

Using Cauchy-Bunyakovsky-Schwarz inequality, we have
$$(x^y + y^x)(1/y + 1/x)
\ge \left(\sqrt{\frac{x^y}{y}} + \sqrt{\frac{y^x}{x}}\right)^2.$$
It suffices to prove that
$$\sqrt{\frac{x^y}{y}} + \sqrt{\frac{y^x}{x}} \ge 2. \tag{1}$$
This inequality has been proved
in show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$.

Remark 0: I have just noticed that @Erik Satie gave an alternative proof for (1) which is simpler than mine. See: show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$
Remark 1: In the link (Chinese website, provided by @mengdie1982) https://www.zhihu.com/question/489824366,
qiqi1509 gave another proof for (1).
The sketch of qiqi1509's proof:
(1) is written as
$$\frac{x}{x + y}x^{(y - 1)/2} + \frac{y}{x + y} y^{(x - 1)/2}
\ge \frac{2\sqrt{xy}}{x + y}.$$
Since $u\mapsto \mathrm{e}^u$ is convex, using Jensen's inequality, we have
$$\frac{x}{x + y}x^{(y - 1)/2} + \frac{y}{x + y} y^{(x - 1)/2}
\ge \mathrm{e}^{\frac{x}{x + y}\frac{y - 1}{2}\ln x
+ \frac{y}{x + y}\frac{x - 1}{2}\ln y}.$$
It suffices to prove that
$$\frac{x}{x + y}\frac{y - 1}{2}\ln x
+ \frac{y}{x + y}\frac{x - 1}{2}\ln y
\ge \ln \frac{2\sqrt{xy}}{x + y}. \tag{2}$$
$\phantom{2}$
Remark 2: (2) is written as
$$\ln \frac{x + y}{2} \ge \frac{2x + (1 - x)y}{2x + 2y}\ln x
+ \frac{x + (2 - x)y}{2x + 2y}\ln y. \tag{3}$$
Perhaps, there is a simpler (or alternative) proof for (3) (with $0 < x \le 1 \le y$).
$\phantom{2}$
Remark 3: A similar inequality is
$$\ln \frac{x + y}{2} \ge \frac{x + 1}{2x + 2y}\ln x
+ \frac{y + 1}{2x + 2y}\ln y. \tag{4}$$
See: Prove $2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0$
I think qiqi1509's idea works for it.
