# 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 ?

Regards Max .

• It could be easier to prove that $x^k+x^{\frac 1 k}+(1-x)^k+(1-x)^{\frac 1 k} \le 2$ for $x \in [0.25,0.75]$ and $k\in[0,1]$ – Bastien Tourand Aug 25 at 18:38
• Your function is not defined on all the interval $\left[\frac{1}{4},\frac{3}{4}\right]$ indeed $\cos(x50\pi)=0$ for $x=\frac{2k+1}{100}$ and $12\le k\le 37$ – Angelo Aug 25 at 21:02
• @Angelo thanks to point out that ! – Erik Satie Aug 26 at 9:46
• @BastienTourand it's more complicated than this . – Erik Satie Aug 26 at 9:46
• Am I the only one for whom some of the formulas appear with unreadably overlapping symbols? For example, the inequality symbol in formula (2) in the "update" section is obscured, for me, by the "$\pi)^2$ part of the exponent. I don't see any errors in the TeX commands, so maybe it's just my browser messing things up. – Barry Cipra Aug 28 at 16:07

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.

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.