Show that : $f(x)+f(1-x)\leq 2$

Question

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 .

Answer 1

There is a partial answer.

Let $y=x-\frac12,$ then $y\in\left(-\frac14,\frac14\right),$ $$f(x)+f(1-x) = f_1(y)+f_2(y),\tag1$$ where $$f_1(y) = \left(\frac12-y\right)^{\large\cos^2(50\pi y)} + \left(\frac12 +y\right)^{\large \cos^2(50\pi y)},\tag{1a}$$ $$f_2(y) = \left(\frac12-y\right)^{\large \sec^2(50\pi y)} + \left(\frac12 +y\right)^{\large \sec^2(50\pi y)}\tag{1b}$$ are the even functions and WLOG $y\in[0,\frac14).$

Taylor series of the function $f_1(y+z_i)+f_2(y+z_i),$ where $z_i = 0.02 (0,1,\dots,12),$ are shown in the table below.

\begin{vmatrix} z_I & f_1(y+z)+f_2(y+z)\\ 0.24 & 2 - 2.08014×10^7 y^4 + 8.6083×10^8 y^5 - 1.6974×10^{11} y^6 + 7.07743×10^{12} y^7\\ & + 2.45289×10^{15} y^8 - 1.40341×10^{17} y^9 + 5.11009×10^{19} y^{10}\\ 0.22 & 2 - 3.74604×10^7 y^4 + 8.04147×10^8 y^5 - 3.06616×10^{11} y^6 + 6.61178×10^{12} y^7\\ & + 5.13588×10^{15} y^8 - 1.2783×10^{17} y^9 + 1.04211×10^{20} y^{10}\\ 0.20 & 2 - 5.29341×10^7 y^4 + 7.42476×10^8 y^5 - 4.3377×10^{11} y^6 + 6.10498×10^{12} y^7\\ & + 7.56646×10^{15} y^8 - 1.15163×10^{17} y^9 + 1.5254×10^{20} y^{10}\\ 0.18 & 2 - 6.71331×10^7 y^4 + 6.76835×10^8 y^5 - 5.50464×10^{11} y^6 + 5.56543×10^{12} y^7\\ & + 9.74392×10^{15} y^8 - 1.02558×10^{17} y^9 + 1.96023×10^{20} y^{10}\\ 0.16 & 2 - 7.99863×10^7 y^4 + 6.08019×10^8 y^5 - 6.56107×10^{11} y^6 + 4.9997×10^{12} y^7\\ & + 1.1671×10^{16} y^8 - 9.015×10^{16} y^9 + 2.34665×10^{20} y^{10}\\ 0.14 & 2 - 9.14369×10^7 y^4 + 5.36659×10^8 y^5 - 7.50227×10^{11} y^6 + 4.413×10^{12} y^7\\ & + 1.33524×10^{16} y^8 - 7.80091×10^{16} y^9 + 2.68512×10^{20} y^{10}\\ 0.12 & 2 - 1.01439×10^8 y^4 + 4.63268×10^8 y^5 - 8.32447×10^{11} y^6 + 3.80956×10^{12} y^7\\ & + 1.47939×10^{16} y^8 - 6.61642×10^{16} y^9 + 2.9763×10^{20} y^{10}\\ 0.10 & 2 - 1.09957×10^8 y^4 + 3.88266×10^8 y^5 - 9.02467×10^{11} y^6 + 3.19284×10^{12} y^7\\ & + 1.60014×10^{16} y^8 - 5.46137×10^{16} y^9 + 3.22099×10^{20} y^{10}\\ 0.08 & 2 - 1.16961×10^8 y^4 + 3.12008×10^8 y^5 - 9.6005×10^{11} y^6 + 2.56577×10^{12} y^7\\ & + 1.69806×10^{16} y^8 - 4.33337×10^{16} y^9 + 3.41994×10^{20} y^{10}\\ 0.06 & 2 - 1.22431×10^8 y^4 + 2.34796×10^8 y^5 - 1.00501×10^{11} y^6 + 1.93084×10^{12} y^7\\ & + 1.77365×10^{16} y^8 - 3.22855×10^{16} y^9 + 3.57388×10^{20} y^{10}\\ 0.04 & 2 - 1.26349×10^8 y^4 + 1.56899×10^8 y^5 - 1.03722×10^{12} y^6 + 1.29026×10^{12} y^7\\ & + 1.82734×10^{16} y^8 - 2.14191×10^{16}{16} y^9 + 3.68338×10^{20} y^{10}\\ 0.02 & 2 - 1.28704×10^8 y^4 + 7.85587×10^7 y^5 - 1.05658×10^{12} y^6 + 6.46029×10^{11} y^7\\ & + 1.85943×10^{16} y^8 - 1.06778×10^{16} y^9 + 3.7489×10^{20} y^{10}\\ 0.00 & 2 - 1.2949×10^8 y^4 - 1.06304×10^{12} y^6 + 1.8701×10^{16} y^8 + 3.77071×10^{20} y^{10} \end{vmatrix}

The plots for $z=0,\, 0.02$

and for $0.24,\ 0.22$

confirm that obtained series correspond to maxima $2$ at $y=0.$

Therefore, $f_1(z_I)+f_2(z_I) = 2$ are the maxima.

Answer 2

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.