How to prove that $x^{1-x}+(1-x)^{x}\leq x^{1/2}+(1-x)^{1/2}$? Let $x\in [0,1]$,try to prove that:
$$x^{1-x}+(1-x)^{x}\leq x^{1/2}+(1-x)^{1/2}$$
My try:

let $x=\sin ^{2}t$,and it is equal to show that  $$\sin^{2}t^{\cos^2{t}}+\cos^{2}t^{\sin^2{t}}\leq \sin t+\cos t$$

but still nothing.

Thanks :-)
I use https://eudml.org/doc/223938 , pdf-file 
page $135$, section $7$ with $\enspace a:=1-x\enspace $ and $\enspace b:=x$ .

Because of the symmetry of the inequality $\enspace x^{1-x}+(1-x)^x\leq \sqrt{x}+\sqrt{1-x}\enspace $ at $\enspace x=0.5$ , 

see $\enspace x\to 1-x$, it’s enough to check the mentioned inequality for $\enspace 0\leq x\leq 0.5$ 

instead of  $\enspace 0\leq x\leq 1$ .

We have to proof for $\enspace 0\leq x\leq 0.5$ :    

$(1.1) \hspace{1cm}\displaystyle x^{1-x}\leq\sqrt{4x^2(1-x)+2x(1-x)(1-2x)\ln(1-x)}$

$(1.2) \hspace{1cm}\displaystyle \sqrt{4x^2(1-x)+2x(1-x)(1-2x)\ln(1-x)}\leq\sqrt{x}(1-(1-x)(1-2x)\ln 2)$

$(1.3) \hspace{1cm}\displaystyle -4+8x^2+(8x-6)(1-x)(1-2x)\ln 2&lt;4x-2+(8x-6)\ln(2(1-x))$

$(1.4) \hspace{1cm}\displaystyle x(3-2x)(3-4x)\ln 2&lt;1$

$(2.1) \hspace{1cm}\displaystyle (1-x)^x\leq\sqrt{1-4x^2(1-x)-2x(1-x)(1-2x)\ln(1-x)}$

$(2.2) \hspace{1cm}\displaystyle \sqrt{1-4x^2(1-x)-2x(1-x)(1-2x)\ln(1-x)}\leq\sqrt{1-x}(1+x(1-2x)\ln 2)$

$(2.3) \hspace{1cm}\displaystyle (1-4x)(\ln 2)^2-\frac{2}{1-x}&lt;-1\leq\frac{3-4x}{(1-x)^2(1-2x)^2}-\frac{4}{1-2x}$

$(3) \hspace{1.3cm}\displaystyle \sqrt{x}(1-(1-x)(1-2x)\ln 2)+\sqrt{1-x}(1+x(1-2x)\ln 2)\leq \sqrt{x}+\sqrt{1-x}$

The proofs are carried out from the bottom upwards.

To $(3)$ :  

$0\leq x\leq 0.5\enspace $ => $\enspace 0\leq x\leq 1-x\enspace $ => $\enspace \sqrt{x}\leq\sqrt{1-x}\enspace $ => $\enspace x\sqrt{1-x}\leq (1-x)\sqrt{x}$

=> $\enspace x\sqrt{1-x}(1-2x)\ln 2\leq (1-x)\sqrt{x}(1-2x)\ln 2$ 

=> $\enspace \sqrt{x}(-(1-x)(1-2x)\ln 2)+\sqrt{1-x}(x(1-2x)\ln 2)\leq 0$  

=> $\enspace \sqrt{x}(1-(1-x)(1-2x)\ln 2)+\sqrt{1-x}(1+x(1-2x)\ln 2)\leq \sqrt{x}+\sqrt{1-x}$

To $(2.3)$ : 

On the one hand it’s $\enspace (1-x)(1-4x)(\ln 2)^2\leq (\ln 2)^2&lt;1\leq 1+x\enspace $ and therefore

$\displaystyle (1-4x)(\ln 2)^2-
\frac{2}{1-x}&lt;-1\enspace $ and on the other hand because of $\enspace x\leq 0.5\enspace $ we get 

$(a)\hspace{1cm}x^2(-1+2x)\leq 0\enspace$ => $\enspace (1-x)^2(3+2x)\leq 3-4x\enspace\enspace$ and 

$(b)\hspace{1cm}(1-x)^2(1-2x)(3+2x)\leq (1-x)^2(3+2x)$ . 

With $\enspace (a) \enspace $ and $\enspace (b) \enspace $ follows $\enspace (1-x)^2(1-2x)(3+2x)\leq 3-4x$ 

and therefore $\enspace\displaystyle -1\leq \frac{3-4x}{(1-x)^2(1-2x)^2}-\frac{4}{1-2x}$ .

To $(2.2)$ based on $(2.3)$ : 

It’s $\enspace\displaystyle \frac{d}{dx}(-2+x(1-2x)(\ln 2)^2+2\ln(2(1-x)))=(1-4x)(\ln 2)^2-\frac{2}{1-x}\enspace$ and 

$\displaystyle \frac{d}{dx}(\frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x})=\frac{3-4x}{(1-x)^2(1-2x)^2}-\frac{4}{1-2x}\enspace$ and it follows 

$\displaystyle \frac{d}{dx}(-2+x(1-2x)(\ln 2)^2+2\ln(2(1-x)))&lt;\frac{d}{dx}(\frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x})$ .

Together with the common point 

$\displaystyle (-2+x(1-2x)(\ln 2)^2+2\ln(2(1-x)))|_{x=0.5}=-2=(\frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x})|_{x=0.5}\enspace $ does it mean that 

$-2+x(1-2x)(\ln 2)^2+2\ln(2(1-x)) \enspace $ and $\enspace\displaystyle \frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x}\enspace $ don’t touch each other for $\enspace 0\leq x&lt;0.5$ .

Taking a value of this value range, e.g. $x=0$, we get  

$\displaystyle (\frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x})|_{x=0}=-1&lt;$

$\displaystyle &lt;-2+2\ln 2=(-2+x(1-2x)(\ln 2)^2+2\ln(2(1-x)))|_{x=0}$ 

and therefore $\enspace\displaystyle \frac{1}{(1-x)(1-2x)}-\frac{2}{1-2x}\leq -2+x(1-2x)(\ln 2)^2+2\ln(2(1-x))$ .

Elementary transformations lead to 

$1-4x^2(1-x)-2x(1-x)(1-2x)\ln(1-x)\leq (1-x)(1+x(1-2x)\ln 2)^2$ .

To $(2.1)$ : $\enspace$ That’s formula $(7.2)$, see the link above.

To $(1.4)$ : 

With $\enspace\displaystyle \frac{d}{dx}(x(3-2x)(3-4x)\ln 2)=((x-\frac{3}{4})^2-\frac{3}{16})24\ln 2:=0\enspace $ follows 

$\displaystyle x_{1,2}=\frac{3\pm\sqrt{3}}{4}\enspace $ and with $\enspace\displaystyle \frac{d^2}{dx^2}(x(3-2x)(3-4x)\ln 2)=(2x-\frac{3}{2})24\ln 2&lt;0$ 

for $\enspace 0\leq x\leq 0.5\enspace $ follows concavity of $\enspace x(3-2x)(3-4x)\ln 2$ . 

Therefore we get 

$ x(3-2x)(3-4x)\ln 2\leq\max(x(3-2x)(3-4x)\ln 2)=$

$=(x(3-2x)(3-4x)\ln 2)|_{x=(3-\sqrt{3})/4}=0.75\sqrt{3}\ln 2&lt;1$ .  

To $(1.3)$ based on $(1.4)$ : 

It’s $\enspace 2x&lt;2x+1-x(3-2x)(3-4x)\ln 2\enspace $ and with

$4x^2\leq 2x\enspace $ we get $\enspace 4x^2&lt;2x+1-x(3-2x)(3-4x)\ln 2$ .

Elementary transformations lead to 

$4x^2+(4x-3)(1-x)(1-2x)\ln 2&lt;2x+1+(4x-3)\ln 2$ .

Because of $\enspace 4x-3&lt;0\enspace $ and $\enspace \ln(2(1-x))\leq \ln 2$ 

we have $(4x-3)\ln 2\leq (4x-3)\ln(2(1-x))\enspace$ and it follows 

$4x^2+(4x-3)(1-x)(1-2x)\ln 2&lt;2x+1+(4x-3)\ln 2\leq$

$\leq 2x+1+(4x-3)\ln(2(1-x))$ 

and with elementary transformations using the left and right side we get 

$-4+8x^2+(8x-6)(1-x)(1-2x)\ln 2&lt;4x-2+(8x-6)\ln(2(1-x))$ .

To $(1.2)$ based on $(1.3)$ : 

It’s 

$\displaystyle \frac{d}{dx}(1-4x(1-x)+((1-x)(1-2x)\ln 2)^2)=-4+8x^2+(8x-6)(1-x)(1-2x)\ln 2$ 

and 

$\displaystyle \frac{d}{dx}(2(1-x)(1-2x)\ln(2(1-x)))=4x-2+(8x-6)\ln(2(1-x))$ 

so that we get 

$\displaystyle \frac{d}{dx}(1-4x(1-x)+((1-x)(1-2x)\ln 2)^2)&lt;\frac{d}{dx}(2(1-x)(1-2x)\ln(2(1-x)))$ . 

Together with the common point  

$(1-4x(1-x)+((1-x)(1-2x)\ln 2)^2)|_{x=0.5}=0=(2(1-x)(1-2x)\ln(2(1-x)))|_{x=0.5}\enspace $ 

does it mean that 

$1-4x(1-x)+((1-x)(1-2x)\ln 2)^2\enspace $ and $\enspace 2(1-x)(1-2x)\ln(2(1-x))$ 

don’t touch each other for $\enspace 0\leq x&lt;0.5$ . 

Taking a value of this value range, e.g.  $x=0$, we get 

$(1-4x(1-x)+((1-x)(1-2x)\ln 2)^2)|_{x=0}=1+(\ln 2)^2&gt;$

$&gt;2\ln 2=(2(1-x)(1-2x)\ln(2(1-x)))|_{x=0}$ 

and therefore $\enspace 2(1-x)(1-2x)\ln(2(1-x))\leq 1-4x(1-x)+((1-x)(1-2x)\ln 2)^2$. 

Elementary transformations lead to 

$4x^2(1-x)+2x(1-x)(1-2x)\ln(1-x)\leq x(1-(1-x)(1-2x)\ln 2)^2$ .

To $(1.1)$: $\enspace$ That’s formula $(7.1)$, see the link above.

Now we have 

$x^{1-x}+(1-x)^x$

$\leq\sqrt{4x^2(1-x)+2x(1-x)(1-2x)\ln(1-x)}$

$\hspace{0.5cm}+\sqrt{1-4x^2(1-x)-2x(1-x)(1-2x)\ln(1-x)}$ 

$\leq\sqrt{x}(1-(1-x)(1-2x)\ln 2)+\sqrt{1-x}(1+x(1-2x)\ln 2)$

$\leq\sqrt{x}+\sqrt{1-x}$ .
Hint:as we see $x+(1-x)=1$ take $a=x ,b=1-x \to a+b=1 , 0&lt; a,b &lt;1$
you have to prove $$a^b+b^a \leq \sqrt a+\sqrt b$$