Proof that if $x,y>0$ and $x+y=1$, then $(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$ For positive reals $x$ and $y$ such that $x+y=1$, prove that $$(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$$
I have tried using Jensen’s inequality but it won’t cover all the possible choices for $x$ and $y$ since the concavity varies. I am trying to find a neat solution so that a generalization could also be made. Thank you.
 A: $$(2x)^{\frac 1 x}=\frac{1}{\left( \frac{1}{2x}\right) ^{\frac{1}{x}}}\leq \frac{1}{1+\frac{1}{x}\left( \frac{1}{2x}-1\right)}=\frac{2x^2}{2x^2-2x+1}$$ by Bernoulli’s inequality. The same holds for $y$ and one immediately computes that if $x+y=1$, $$\frac{2x^2}{2x^2-2x+1}+\frac{2y^2}{2y^2-2y+1}=2$$ and the result follows.(just observe that the denominator are the same $=x^2+y^2$)
A: Question : Can we use Jensen's inequality here ? No .
Hint :
One can show that the function :
$$f(x)=1-\frac{1}{(2a(1-x)+2bx)^{\frac{1}{a(1-x)+bx}}}$$
Is concave on $I=[0,1]$ where $1>a\geq 0.5\geq b>0$ and $a+b=1$
So we can apply Jensen's inequality with weight we get :
$$(2a)^{\frac{1}{a}}f(0)+(2b)^{\frac{1}{b}}f(1)\leq ((2a)^{\frac{1}{a}}+(2b)^{\frac{1}{b}})f\Bigg(\frac{(2b)^{\frac{1}{b}}}{(2a)^{\frac{1}{a}}+(2b)^{\frac{1}{b}}}\Bigg)$$
But :
$$\frac{(2b)^{\frac{1}{b}}}{(2a)^{\frac{1}{a}}+(2b)^{\frac{1}{b}}}\leq 0.5$$
One can show that the function $f(x)$ is decreasing on $I$ so :
$$f\Bigg(\frac{(2b)^{\frac{1}{b}}}{(2a)^{\frac{1}{a}}+(2b)^{\frac{1}{b}}}\Bigg)\geq f(0.5)=0$$
Now we can conclude !
Update :
I think we can use weighted Karamata's inequality (but I'm not sure) to get :
$$(2a)^{\frac{1}{a}}f(0)+(2b)^{\frac{1}{b}}f(1)\leq (2a)^{\frac{1}{a}}f(0.5)+(2b)^{\frac{1}{b}}f(0.5)=0$$
Since :$$0(2a)^{\frac{1}{a}}\leq 0.5(2a)^{\frac{1}{a}} \quad 0.5(2a)^{\frac{1}{a}}+0.5(2b)^{\frac{1}{b}}\geq 1(2b)^{\frac{1}{b}}+0(2a)^{\frac{1}{a}}$$
Wich is the desired inequality .
I think that the update is false but with strong convexity it seems to work (numerically) with particular value for $a,b$ .
Curious fact :
We cannot use Jensen's inequality directly but we can use it to refine the result and I found it very strange .
One can show that the function :
$$g(x)=\frac{1}{\ln\Big((2x)^{\frac{1}{x}}+(2(1-x))^{\frac{1}{1-x}}-1\Big)}$$
where $0<x<0.5$ is concave .To prove it we can use the proof of William Sun .
Now except the extrema and the equality case the equality :
$$h(x)=(2x)^{\frac{1}{x}}+(2(1-x))^{\frac{1}{1-x}}=c\quad (1)$$
Have four solutions on $I=(0,1)$
So two solutions on $0<x<0.5$ and two solutions on $0.5<x<1$ we can express it as :
$$h(x)=h(y)=h(1-y)=h(1-x)$$
Where $0<x<y<0.5$
So if we apply Jensen inequality we have :
$$g(x)+g(y)=g(x)+g(1-x)=2 g(x)\leq 2g\Big(\frac{x+y}{2}\Big)$$
To conclude we can use numerical methods to solve $(1)$
