Show that : $f(x)+f(1-x)\leq 2$ I'm very proud to show one of my dream in term of inequalities .
Claim

Let $0.25\leq x\leq 0.75$ and $x\neq \frac{2k+1}{100}$ with $12\leq k\leq 37$ and $k$ a natural number then define the function :
$$f(x)=x^{\frac{1}{\cos^2(x50\pi)}}+x^{\cos^2(x50\pi)}$$
then we have :
$$f(x)+f(1-x)\leq 2$$

First we have $50$ (limit) equality cases as $x=\frac{25}{100},\frac{26}{100},\frac{27}{100},\cdots,\frac{73}{100},\frac{74}{100},\frac{75}{100}$
To prove it I have tried Bernoulli's inequality as we have :
$$x^{\frac{1}{\cos^2(x50\pi)}}\leq \frac{1}{1+\Big(\frac{1}{x}-1\Big)\frac{1}{\cos^2(x50\pi)}}$$
And :
$$x^{\cos^2(x50\pi)}\leq 1+(x-1)\cos^2(x50\pi)$$
But it doesn't work .
I add a graph to convince you :

Update as partial answer :
It's an heavy method but it works numerically speaking . Well we show that the inequality is true for $x\in[0.307,0.31)$ and $x\in(0.31,0.313]$ . Firstly on these intervals we have :
$$(1-x)^{\cos((1-x)50\pi)^2}+x^{\frac{1}{\cos(x50\pi)^2}}\leq 1\quad (1)$$
And
$$x^{\cos(x50\pi)^2}+(1-x)^{\frac{1}{\cos((1-x)50\pi)^2}}\leq 1\quad(2)$$
Now we use the method used here General trick to factorize an inequality of the kind $a+b\leq 1$ . The problem becomes :
$$\sin\Big(x^{\frac{1}{\cos(x50\pi)^2}}\frac{\pi}{2}\Big)\leq \cos\Big((1-x)^{\cos((1-x)50\pi)^2}\frac{\pi}{2}\Big)$$
Or :
$$\ln\Big(x^{\frac{1}{\cos(x50\pi)^2}}\frac{\pi}{2}\Big)\leq \ln \Big(\sin^{-1}\Big(\cos\Big((1-x)^{\cos((1-x)50\pi)^2}\frac{\pi}{2}\Big)\Big)\Big)$$
We study the function :
$$h(x)= \ln \Big(\sin^{-1}\Big(\cos\Big((1-x)^{\cos((1-x)50\pi)^2}\frac{\pi}{2}\Big)\Big)\Big)-\ln\Big(x^{\frac{1}{\cos(x50\pi)^2}}\frac{\pi}{2}\Big)$$
The derivative is here
Studing this function we see that for $x\in[0.307,0.31)$ the function is increasing and decreasing for $x\in(0.31,0.313]$
But :
$$f(0.307)>0 \quad \operatorname{and} \quad f(0.313)>0$$
Happy ending !
Question
How to show my claim ?
Thanks in advance !
Regards Max .
 A: Note $\cos(50\pi (1-x)) = \cos(50\pi x)$ so indeed if we can prove that $f(x,k) = x^k+x^{\frac 1 k}+(1-x)^k+(1-x)^{\frac 1 k} \le 2$  for $ x \in [0.25,0.75]$ and $k \in (0, 1]$, we have proved a more general result than asked here.
If we fix $x$ and inspect $f(x,k)$ as a function of $k$ then it shows that for all $x$,  $f(k)$ has only one minimum w.r.t. $k$, and the behavior is that  $f(k=0) \to 2$, then $f(k)$ is falling monotonously with $k$ towards that minimum (interval 1), then    $f(k)$ is rising monotonously (interval 2) until  it reaches $f(k=1) = 2$.
To show this in the two intervals defined above, look at the derivatives. We have
$$
\partial f(x,k) / \partial k  = \log(x) [x^k-\frac{1}{k^2}x^{\frac 1 k}] + \log(1-x) [ (1-x)^k-\frac{1}{k^2}(1-x)^{\frac 1 k} ]
$$
Consider interval 1. (The proof is yet given for this part.)
The two terms $x^k$ and $(1-x)^k$ are falling with $k$. So for establishing that no further solution $\partial f(x,k) / \partial k  = 0$ exists, it is enough if we can show that also the terms  ${k^2}x^{- \frac 1 k}$ and ${k^2}(1-x)^{-\frac 1 k}$ are falling with $k$. Let's again show this with calculus. Setting $g(k) = {k^2}x^{- \frac 1 k}$ gives $g'(k) = (2{k} + \log(x)) x^{- \frac 1 k}$ which is negative as long as $ k< - \frac12 \log(x)$. Likewise for the other term we require $ k< - \frac12 \log(1-x)$. Since we are in interval 1, we have (by inspection of the minimum which is at $x <0.5$) that the relevant (harder) condition is $ k< - \frac12 \log(1-x)$. However, this regime is actually larger than the required regime for interval 1, which can be seen by evaluating $\partial f(x,k) / \partial k $ at the limit $ k= - \frac12 \log(1-x)$ which shows that $\partial f(x,k) / \partial k > 0 $ for all $x$. This means that the condition $ k< - \frac12 \log(1-x)$ actually reaches into interval 2 where $f(k)$ is rising again, and we are safe. This proves interval 1.
Interval 2 should be proven similarly, I just didn't find time to do that yet.
