Can we find a necessary and sufficient condition to have : $(x^2)^{1-f(x)}+((1-x)^2)^{1-f(1-x)}\leq 1$? It's a possible generalization of an inequality of Vasile Cirtoaje I recall :


Let $x,y>0$ such that $x+y=1$ then we have :
$$x^{2y}+y^{2x}\leq 1$$


Well the idea is really simple take a function $f(x)$ continuous an differentiable on $I=(0,1)$ such that $f(0)=0$ and $f(0.5)=0.5$ and finally $f(1)=1$ .What conditions make to have :
$$(x^2)^{1-f(x)}+((1-x)^2)^{1-f(1-x)}\leq 1\quad (1)$$
We can put :
$$f(x)=0.5\frac{g(x)}{g(0.5)}$$
Moreover I think it works with increasing functions . Maybe we can add the idea of convexity (?).
I have tested example as $g(x)=\sin(x)$ or $g(x)=e^x$ and $g(x)=\tanh(x)$ and it seems to work .
My question :
Can we find a necessary and sufficient conditions to have the inequality $(1)$ ?
Thanks in advance !!
Update :
Let $0<x<1$ then define :
$$h(x)=x^{g(x)}$$
We have as constraints :
$1)$ $\lim_{x\to 0^+}h(x)=0$ and $h(0.5)=0.5$ and $\lim_{x\to 1^-}h(x)=1$
$2)$ $(h(x))'\geq 0 $ $\forall x\,\operatorname{such that}\, 0<x<1$
$3)$ $g(x)$ is convexe on $(0,1)$
$4)$ $h(x)$ is continuous on $(0,1)$ and twice differentiable on this interval
Then define the three functions :
$p(x)=h(x)+h(1-x)$ $\quad$ $q(x)=x^{-g(x)}$ and $f(x)=q(x)+q(1-x)$
And the last constraints :
If $p(x)\leq 1$ then the only equality case is $x=0.5$
$f(x)$ is continuous on $(0,1)$ and twice differentiable on this interval
Now my conjecture :
$$\forall x \operatorname{such that} 0<x<1 \,, p(x)\leq 1 \Leftrightarrow f(x)\ln(p(x)) \operatorname{is concave on}\,(0,1)$$
As in my answer we can use Jensen's inequality to prove one of the logical implications .
For the other we can use the definition of the log concavity :
$$(r'(x))^2\geq r''(x)r(x)$$
In the fact there is two big blocks  which cancel each other out .Remains a part of the second derivative  that I cannot study .
Any helps or initiatives would be very appreciated .
 A: Let $\;h(x):\left[0,1\right]\to\mathbb{R}\;$ be the function defined as
$h(x)=\left(x^2\right)^{1-f\left(x\right)}-\frac{1}{2}\;\;\;\;$ for all $\;x\in\left[0,1\right],$
where $f(x):\left[0,1\right]\to\mathbb{R}$ is a function such that
$f(0)=0$, $\;\;\;f\left(\frac{1}{2}\right)=\frac{1}{2}$, $\;\;\;f(1)=1$.
It results that
$h(0)=-\frac{1}{2}$, $\;\;\;h\left(\frac{1}{2}\right)=0$, $\;\;\;h(1)=\frac{1}{2}$,
$h(x)>-\frac{1}{2}\;\;$ for all $\;x\in\left]0,1\right]\;$.
Moreover, the inequality
$$\left(x^2\right)^{1-f(x)}+\left[\left(1-x\right)^2\right]^{1-f(1-x)}\leq 1\quad \color{blue}{(*)}$$
is equivalent to
$$h(x)+h(1-x)\le0.$$
So, if $\;\alpha(x):[0,1]\to\mathbb{R}\;$ is any function such that
$\alpha(1-x)\le -\alpha(x)\;\;\;\forall x\in[0,1]\;,\;$
$\alpha(x)>-\frac{1}{2}\;\;$ for all $\;x\in\left]0,1\right]\;\;$ and
$\alpha(0)=-\frac{1}{2}$, $\;\;\;\alpha\left(\frac{1}{2}\right)=0$, $\;\;\;\alpha(1)=\frac{1}{2}\;$,
then the function
$f(x)= \begin{cases}
  1-\frac{\ln\left[\alpha(x)+\frac{1}{2}\right]}{2\ln x}\;\;\;\;\text{ for all } x\in\left]0,1\right[\\
  0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{ for } x=0\\
  1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{ for } x=1
\end{cases}$
satisfies the inequality $(*)$ for all $x\in\left[0,1\right]$ and
$f(0)=0$, $\;\;\;f\left(\frac{1}{2}\right)=\frac{1}{2}$, $\;\;\;f(1)=1$.
Conversely, if $f(x):\left[0,1\right]\to\mathbb{R}$ is a function which satisfies the inequality $(*)$ for all $x\in\left[0,1\right]$ and
$f(0)=0$, $\;\;\;f\left(\frac{1}{2}\right)=\frac{1}{2}$, $\;\;\;f(1)=1$,
then there exists a function $\;\alpha(x):[0,1]\to\mathbb{R}\;$ such that
$\alpha(1-x)\le -\alpha(x)\;\;\;\forall x\in[0,1]\;,\;$
$\alpha(x)>-\frac{1}{2}\;\;$ for all $\;x\in\left]0,1\right]\;,\;$
$\alpha(0)=-\frac{1}{2}$, $\;\;\;\alpha\left(\frac{1}{2}\right)=0$, $\;\;\;\alpha(1)=\frac{1}{2}\;$ and
$f(x)= \begin{cases}
  1-\frac{\ln\left[\alpha(x)+\frac{1}{2}\right]}{2\ln x}\;\;\;\;\text{ for all } x\in\left]0,1\right[\\
  0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{ for } x=0\\
  1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{ for } x=1
\end{cases}$
