<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] Using again Generalized Young inequality we have :
$$\left(\frac{1}{a^{b}}+\frac{1}{x^{b}}\right)\left(a^{\sqrt{2^{-1}}x^{\left(1-b\right)}}\cdot x^{\left(\sqrt{2^{-1}}\left(a\right)^{\left(1-b\right)}\right)}\cdot a^{\frac{\left(b-0.5\right)\sqrt{2}}{a^{b}}}\cdot x^{\frac{\left(b-0.5\right)\sqrt{2}}{x^{b}}}\right)^{\frac{1}{\sqrt{2}a^{-b}+\frac{\sqrt{2}}{x^{b}}}}\leq \left(\sqrt{\frac{x^{a}}{a}}+\sqrt{\frac{a^{x}}{x}}\right)$$
where $b\to1$ and $a,x>0$
Final conjecture :
Let $a,x>0$ then it seems we have :
$$\left(\frac{1}{a^{b}}+\frac{1}{x^{b}}\right)\left(a^{\sqrt{2^{-1}}x^{\left(1-b\right)}}\cdot x^{\left(\sqrt{2^{-1}}\left(a\right)^{\left(1-b\right)}\right)}\cdot a^{\frac{\left(b-0.5\right)\sqrt{2}}{a^{b}}}\cdot x^{\frac{\left(b-0.5\right)\sqrt{2}}{x^{b}}}\right)^{\frac{1}{\sqrt{2}a^{-b}+\frac{\sqrt{2}}{x^{b}}}}\geq 2$$
Where $b\to 1$
I think we can settle $b=1$ it works also .
Using a bit of algebra (introducing log and invert the variables)
We need to show :
$x,a > 0$, we need to prove that
$$f\left(x\right)=(x+a)\ln\frac{a+x}{2}-\frac{\left(a+1\right)\ln a+\left(x+1\right)\ln x}{2} \ge 0.$$
A proof of this fact can be found here :
Prove $2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0$
</s>