Define the function $1<x<\frac{103}{100}$:
$$f(x)=x^{2 \Bigg(1-0.5 \Bigg(\frac{\Gamma\Big(\frac{1}{x}\Big)}{\Gamma\Big(\frac{1}{2}\Big)}\Bigg)\Bigg)}+(1-x)^{2 \Bigg(1-0.5 \Bigg(\frac{\Gamma\Big(\frac{1}{1-x}\Big)}{\Gamma\Big(\frac{1}{2}\Big)}\Bigg)\Bigg)}$$
Prove that :
$$f(x)<\frac{3}{2}x-\frac{1}{2}$$
I strongly believe that it is a problem of tangent .
So the natural way is to use derivative but first there is a limit that I cannot evaluate
$$\lim_{x\to 1^{+}}(1-x)^{2 \Bigg(1-0.5 \Bigg(\frac{\Gamma\Big(\frac{1}{1-x}\Big)}{\Gamma\Big(\frac{1}{2}\Big)}\Bigg)\Bigg)}=0$$
I have tried power series at the first order of the Gamma function ($x=1$) we have :
$$\Gamma\Big(\frac{1}{1-x}\Big)=1-\gamma\Big(\frac{1}{1-x}-1\Big)+O\Big(\Big(\frac{1}{1-x}-1\Big)^2\Big)$$
But it's ineffective unfortunately .
Why this limit ?
Because it implies that if:
$$g(x)=(1-x)^{2 \Bigg(1-0.5 \Bigg(\frac{\Gamma\Big(\frac{1}{1-x}\Big)}{\Gamma\Big(\frac{1}{2}\Big)}\Bigg)\Bigg)}$$
Then :
$$\lim_{x\to 1^{+}}g'(x)=0$$
Then we have just to evaluate the derivative of :
$$h(x)=x^{2 \Bigg(1-0.5 \Bigg(\frac{\Gamma\Big(\frac{1}{x}\Big)}{\Gamma\Big(\frac{1}{2}\Big)}\Bigg)\Bigg)}$$
As $x\to 1^{+}$
An conclude with the formula :
$$y=f'(x_0)(x-x_0)+f(x_0)$$
My question :
How to prove the strict inequality ?
Thanks in advance for your help !