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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 :-)

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    $\begingroup$ Note :$$\sin^{2}t^{\cos^2{t}}+\cos^{2}t^{\sin^2{t}}\leq |\sin t|+|\cos t|$$ $\endgroup$ – Khosrotash Mar 5 '17 at 14:58
  • $\begingroup$ I don't know how to show it ,can you give me some hints? $\endgroup$ – Schröchin Mar 5 '17 at 15:02
  • $\begingroup$ Easy to show that $(1-x)^x+x^{1-x}\leq\sqrt2$ $\endgroup$ – Michael Rozenberg Mar 5 '17 at 17:29
  • $\begingroup$ You could use complete induction as inequality is true for base case i.e. 0 and assume true for $x < 1$ .Since, it also holds for 1... It will definitely hold in the specified interval.... :) $\endgroup$ – user35508 Mar 6 '17 at 14:28
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    $\begingroup$ @Schröchin: Evaluate $a$ and $b$ with $ax+b$ is a tangent for $((1-x)^{x-\frac{1}{2}}-1)x^{-\frac{1}{2}}$ in $x=\frac{1}{2}$ and show that it is also a tangent for $(1-x^{\frac{1}{2}-x})(1-x)^{-\frac{1}{2}}$ . Then it's not difficult any more to show $((1-x)^{x-\frac{1}{2}}-1)x^{-\frac{1}{2}} \leq ax+b \leq (1-x^{\frac{1}{2}-x})(1-x)^{-\frac{1}{2}}$ . I hope it works. :-) $\endgroup$ – user90369 Mar 6 '17 at 16:35
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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 $\enspace 0\leq x\leq 0.5$ :

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

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

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

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

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

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

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

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

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<1\leq 1+x\enspace $ and therefore

$\displaystyle (1-4x)(\ln 2)^2- \frac{2}{1-x}<-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

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

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 $\enspace 0\leq x<0.5$ .

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

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

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

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

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)=$

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

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

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

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

Elementary transformations lead to

$4x^2+(4x-3)(1-x)(1-2x)\ln 2<2x+1+(4x-3)\ln 2$ .

Because of $\enspace 4x-3<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

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

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

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

$-4+8x^2+(8x-6)(1-x)(1-2x)\ln 2<4x-2+(8x-6)\ln(2(1-x))$ .

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

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

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 $\enspace 0\leq x<0.5$ .

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

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

$>2\ln 2=(2(1-x)(1-2x)\ln(2(1-x)))|_{x=0}$

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

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

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

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

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

$\leq\sqrt{x}+\sqrt{1-x}$ .

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  • $\begingroup$ thank you very much.nice solution.But I still have a question:how to get \frac{1-x^{\frac{1}{2}-x}}{\frac{1}{2}-x} is convex compared with \frac{a^x-1}{x} $\endgroup$ – Schröchin Apr 15 '17 at 3:03
  • $\begingroup$ @Schröchin : You are right, I've made a mistake, I am very sorry. Please remove the credits to show interested people, that your question isn't still answered. (I have done only a half proof. I cannot conclude directly from $\frac{a^x-1}{x}$ .) $\endgroup$ – user90369 Apr 16 '17 at 6:16
  • $\begingroup$ @Schröchin : The proof is indeed completed now. $\endgroup$ – user90369 May 6 '17 at 8:48
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Hint:as we see $x+(1-x)=1$ take $a=x ,b=1-x \to a+b=1 , 0< a,b <1$ you have to prove $$a^b+b^a \leq \sqrt a+\sqrt b$$

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    $\begingroup$ How does that simplify the task? $\endgroup$ – Martin R Mar 6 '17 at 14:08
  • $\begingroup$ @MartinR Does this help? I think this was what he intended to do, though I may be mistaken. $\endgroup$ – S.C.B. Mar 6 '17 at 14:20

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