show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$

Update: I found a simpler proof.

WLOG, assume that $b \le a$.

If $a, b > 1$ or $a, b < 1$, then $a^{b - 1}\ge 1$ and $b^{a - 1}\ge 1$ and thus $$\sqrt{\frac{a^b}{b}} + \sqrt{\frac{b^a}{a}}\ge 2\sqrt[4]{a^{b-1}b^{a-1}} \ge 2.$$

It remains to prove the case when $0 < b \le 1 \le a$.

Let $a = x^2, b = y^2$. It suffices to prove that, for all $0 < y \le 1 \le x $, $$\frac{x^{y^2}}{y} + \frac{y^{x^2}}{x} \ge 2.$$

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Fact 1: If $x \ge 1$ and $0 < y \le 1$, then $$x^{y^2} \ge \frac{1 + x + (x - 1)y^2}{1 + x - (x - 1)y^2}.$$ (Proof: Let $f(x) = y^2\ln x - \ln \frac{1 + x + (x - 1)y^2}{1 + x - (x - 1)y^2} $. We have $f'(x) = \frac{(1 - y^4)y^2(x - 1)^2}{x[(1 + x)^2 - (x - 1)^2y^4]}\ge 0$. Also, $f(1) = 0$. Thus, $f(x) \ge 0$ for all $x\ge 1$.)

Fact 2: If $x \ge 1$ and $0 < y \le 1$, then $$y^{x^2} \ge \frac{1 + y + (y - 1)x^2}{1 + y - (y - 1)x^2}.$$ (The proof is given at the end.)

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Now, using Facts 1-2, it suffices to prove that $$\frac{1}{y}\cdot \frac{1 + x + (x - 1)y^2}{1 + x - (x - 1)y^2} + \frac{1}{x}\cdot \frac{1 + y + (y - 1)x^2}{1 + y - (y - 1)x^2} \ge 2.$$

Let $x = 1 + s$ for $s \ge 0$. After clearing the denominators, it suffices to prove that $$q_4 s^4 + q_3 s^3 + q_2 s^2 + q_1s + q_0 \ge 0 \tag{1}$$ where \begin{align*} q_4 &= (1 - y)(2y^3 + y^2 - 2y + 1), \\ q_3 &= (1 - y)(7y^3 + 3y^2 - 11y + 5), \\ q_2 &= - 6y^4 + 8y^3 + 24y^2 - 32y + 10, \\ q_1 &= -2y^4 + 4y^3 + 16y^2 - 28y + 10, \\ q_0 &= 4(1 - y)^2. \end{align*} It is easy to prove that $q_4, q_3, q_2, q_0 \ge 0$. Also, we have \begin{align*} 4q_2q_0 - q_1^2 = 4(y^3 + y^2 + 7y + 15)(1 - y)^5 \ge 0. \end{align*} Thus, (1) is true.

We are done.

Proof of Fact 2: We only need to prove the case when $\frac{1 + y + (y - 1)x^2}{1 + y - (y - 1)x^2} > 0$, i.e. $y > \frac{x^2 - 1}{x^2 + 1}$. In other words, we only need to prove the case when $\frac{x^2 - 1}{x^2 + 1} < y \le 1$. Let $$g(y) = x^2\ln y - \ln \frac{1 + y + (y - 1)x^2}{1 + y - (y - 1)x^2}.$$ We have $$g'(y) = - \frac{x^2(x^4 - 1)(1 - y)^2}{y[1 + y + (y - 1)x^2]^2}\cdot \frac{1 + y + (y - 1)x^2}{1 + y - (y - 1)x^2} \le 0.$$ Also, $g(1) = 0$. Thus, $g(y) \ge 0$ for all $y \in (0, 1]$.

We are done.

My try (please point out the errors if any!):

Multiplying both sides by $\sqrt{ab}$, we get that $$\sqrt{a^{b+1}}+\sqrt{b^{a+1}}\geq 2\sqrt{ab}\tag{1}$$ $$\frac{\sqrt{a^{b+1}}+\sqrt{b^{a+1}}}{2}\geq \sqrt{ab}$$ Using AM-GM inequality,

$$\frac{\sqrt{a}+\sqrt{b}}{2}\geq \sqrt{\sqrt{ab}}$$

For $0\leq a\leq b\leq 1$, we have that $\sqrt{\sqrt{ab}}\geq\sqrt{ab}$, $\sqrt{a^{b+1}}\geq \sqrt{a}$ and $\sqrt{b^{a+1}}\geq \sqrt{b}$. Therefore, $$\frac{\sqrt{a^{b+1}}+\sqrt{b^{a+1}}}{2}\geq\frac{\sqrt{a}+\sqrt{b}}{2}\geq \sqrt{\sqrt{ab}}\geq \sqrt{ab}$$

Thus, the original inequality holds.

For $1\leq a\leq b$, we can transform $(1)$ into $$a^{b+1}+b^{a+1}+2{a^{\frac{b+1}{2}}b^{\frac{a+1}{2}}}\ge 4ab$$

As ${a^{\frac{b+1}{2}}b^{\frac{a+1}{2}}}\geq ab$, it follows that $4ab-2{a^{\frac{b+1}{2}}b^{\frac{a+1}{2}}}\leq 2$. As $a^{b+1}+b^{a+1}\geq 2$, the original inequality holds in this case too.

QED