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 $(1)$