<s>[INST] show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$ let $a,b>0.$ Show that
$$\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2\tag{1}$$
I known  How to prove $a^b+b^a>1$,where $a,b>0.$ See $x^y+y^x>1$ for all $(x, y)\in \mathbb{R_+^2}$
to prove $（1）$, I want use AM-GM inequality
$$\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2\left(\dfrac{a^b}{b}\cdot\dfrac{b^a}{a}\right)^{1/4}=2\left(a^{b-1}b^{a-1}\right)^{1/4}$$
But $a^{b-1}b^{a-1}$ is not always  $>1$
 [/INST] 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
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