Prove that :$f(a^2)+f(b^2)+f(c^2)\leq f\left(\frac{1}{4}\right)+f\left(\frac{1}{4}\right)+f(0)$ where $f(x)=\sqrt{\frac{1+\sqrt{1+x}}{x^x}}$

Question

Hi it's a problem found by myself :

Let $0<x<1$ then define :

$$f(x)=\sqrt{\frac{1+\sqrt{1+x}}{x^x}}$$

Then let $a,b,c>0$ such that $a+b+c=1$ then we have :

$$f(a^2)+f(b^2)+f(c^2)\leq f\left(\frac{1}{4}\right)+f\left(\frac{1}{4}\right)+\sqrt{2}\quad(I)$$

I admit that $f(0)$ is a limit case equal to $\sqrt{2}$ .

My strategy use convexity . First difficulty the function $f(x^2)$ is neither convex or concave on $[0,1]$ but the function $f(x)$ is concave on $(0,1)$.So we use Karamata's inequality :

If $a\geq b \geq c $ and if we have :

$$a^2\leq \frac{1}{4}$$

$$a^2+b^2\leq \frac{1}{4}+\frac{1}{4}$$

$$a^2+b^2+c^2= \frac{1}{4}+\frac{1}{4}+0$$

The majorization is complete and we are done for this case .In fact this idea is good but doesn't work since it's concave on $(0,1)$ .So maybe we can work with a small variable (near zero) .

My question :

How to prove $(I)$ ?

Thanks in advance !