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  $>1$
 A: 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
A: Alternative sketch of proof for $0<x\leq 1\leq a$
Using derivative we have the inequalities  :
$$\left(1+\left(a\cdot\frac{1}{1-\left(a-1\right)\left(x-1\right)}-1\right)x\right)\leq a^{x^2}$$
$$\left(1+\left(x\cdot\frac{1}{1-\left(a-1\right)\left(x-1\right)}-1\right)a\right)-x^{a^{2}}\leq 0$$
Then we need to show :
$$\frac{\left(1+\left(a\cdot\frac{1}{1-\left(a-1\right)\left(x-1\right)}-1\right)x\right)}{x}+\frac{\left(1+\left(x\cdot\frac{1}{1-\left(x-1\right)\left(a-1\right)}-1\right)a\right)}{a}\geq 2$$
Wich is easy !
A: 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 $2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0$
A: COMMENT.- I have trouble writing in English so this schematic presentation.
► $f(x,y)=\sqrt{\dfrac{x^y}{y}}$
► One has to prove for all point $(a,b)$ of the first quadrant
$$f(a,b)+f(b,a)\ge 2$$
► For all positive value $k$ the curves $f(x,y)=k$ and $f(y,x)=k$ are symmetric respect to the line $y=x$.

► One can use the black curve above to reduce the problem to prove for the complement of the region defined by $$f(x,y)\ge 2$$ because for any point in the shadow region the proposed inequality is trivially verified (one term of the sum, $f(x,y)$, is already greater than $2$). Note that a neighborhood of $0$ is discarded which is “natural” because a denominator very small is involved.

► Now we can reduce the white region with the symmetric function $f(y,x)=\sqrt{\dfrac{y^x}{x}}=2$ (note for all point $(b,a)$  “above” this curve its symmetric point $(a,b)$ satisfies $f(a,b)\gt2$ so $f(a,b)+f(b,a)\ge 2$).

► So it remains to prove the inequality for the points inside the white region.
HINT.- Let the line $L: y = -x + a$ where $a$ runs through the interval $[0.36,5.02]$ (these numbers correspond to the values of $a$ for the points of intersection of the two considered symmetric curves). We are going to prove that each point  $(a,b)$  of the segment $\overline{PQ}$, where $P$ and $Q$ are the points of intersection of the line with said curves, satisfies the inequality.

For this, let $F$ be the function
$$F(x)=\sqrt{\frac{x^{a-x}}{a-x}}+\sqrt{\frac{(a-x)^x}{x}}$$ and study its variation. For example, for $a=4$, if the point $P=(x_0,y_0)$ then $x_0\approx 0.426$ and $F(0.426)\approx2.1247\gt2$ (ideally obviously must be equal to $2$). At the point where $x=y$, when $x=2$ (because of $x=-x+4$) one has $F(2)\approx2.8284\gt2$ and the minimum of $F(x)$is equal to $2.087\gt2$ and the study of the curve can be stop here (by symmetry).
Similarly with all other segment with $a\in[0.36,5.02]$ , the problem with $a\ne4$  is analogue. You can verify that for values of  $a$ in $[1.77,2.3]$ the minimum of the function $F(x)$ is very near of  $2$ but always greater that $2$.
It is clear that you can also calculate the minimum of $F(x)$ for a literal $a$ belonging to the interval above mentioned.

