Show that $\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\cos\left(\frac{2}{x}\right)>0$ for $x\geq 4/\pi$

Question

It's a follow up of Proving that ${(1-\frac{2}{x^2})}^x < \frac{x-1}{x+1}$ for any $x > 2$. :

Let $x>4/\pi$ then we have :

$$\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\cos\left(\frac{2}{x}\right)>0$$

My strategy :

For $x\geq 2$ we have :

$$\cos\left(\frac{2}{x}\right)-\cos\left(\frac{2}{\sqrt{x}}\right)\geq 2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|\geq \left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)^{\frac{1}{\sqrt{x}}}$$

Remains to show :

$$\lim_{n\to \infty}\cos\left(\frac{2}{x^{2n}}\right)-\left(\frac{x^{2n}-1}{x^{2n}+1}\right)^{\frac{1}{x^{2n}}}=0$$

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Lemma 1 :

Let $x\geq 1$ $$\frac{\left(x^{2}-1\right)}{x^{2}+1}-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}$$

Proof :

For the RHS we have (using the same approach as in my answer):

$$f(x)=1+\frac{1}{x}\left(\frac{x-1}{x+1}-1\right)+\frac{1}{2}\left(\frac{1}{x}-1\right)\frac{1}{x}\left(\frac{x-1}{x+1}-1\right)^{2}+\frac{1}{6}\left(\frac{1}{x}-1\right)\frac{1}{x}\left(\frac{1}{x}-2\right)\left(\frac{x-1}{x+1}-1\right)^{3}\geq\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}$$

And :

$$f(x)-\frac{x^{2}-1}{x^{2}+1}=-\frac{2\left(x-1\right)\left(x^{3}-2x^{2}+7x-2\right)}{3\left(x^{2}+x\right)^{3}\left(x^{2}+1\right)}\leq 0$$

Now we need to show for $x\geq 2$:

$$\frac{x^{4}-1}{x^{4}+1}-2\frac{x^{2}-1}{x^{4}+\frac{2}{3}+\frac{1}{3}x^{2}}-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}\geq 0$$

But for $x\geq 2$: $$\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}\geq g(x)=\left(\frac{x-1}{x+1}\right)^{-1}\left(1+\left(\frac{1}{x}+1\right)\left(\frac{x-1}{x+1}-1\right)+\frac{1}{2}\left(\frac{1}{x}+1\right)\frac{1}{x}\left(\frac{x-1}{x+1}-1\right)^{2}+\frac{1}{6}\left(\frac{1}{x}-1\right)\frac{1}{x}\left(\frac{1}{x}+1\right)\left(\frac{x-1}{x+1}-1\right)^{3}\right)$$

But :

$$\frac{x^{4}-1}{x^{4}+1}-2\frac{x^{2}-1}{x^{4}+\frac{2}{3}+\frac{1}{3}x^{2}}-g(x)=\frac{-2\left(3x^{8}-3x^{7}-x^{6}-3x^{5}+4x^{4}-6x^{3}-4x^{2}-6x+4\right)}{(3x^{3}(x+1)(x^{4}+1)(3x^{4}+x^{2}+2))}<0$$

So the RHS is shown for $x\ge 2 $

Now the LHS :

$$h(x)=\cos\left(\frac{2}{x}\right)-\cos\left(\frac{2}{\sqrt{x}}\right)-2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|$$

We have for $x\in[0,\pi/2]$:

$$1-\frac{x^{2}}{2}+\frac{x^{4}}{24}\geq \cos(x)\geq 1-\frac{x^{2}}{2}$$

So for $x\geq 4 $ we have :

$$r\left(\frac{2}{x}\right)-t\left(\frac{2}{\sqrt{x}}\right)-2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|\leq 0$$

Where :

$$t\left(x\right)=1-\frac{x^{2}}{2}+\frac{x^{4}}{24},r\left(x\right)=1-\frac{x^{2}}{2}$$

So we are done for $x\geq 4$.

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Have you another (simpler) proof ?

Bonus/conjecture :

Let $x>1$ such that $x\in[\alpha,\beta]$ then define :

$$m\left(x\right)=\cos\left(\frac{2}{x}\right)-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}$$

Then it seems $\exists a,b,c\in[-\infty,\infty]$ and $0<\varepsilon$ arbitrary small such that :

$$b^{-x^{a}}+c-\varepsilon<\sum_{n=1}^{\infty}\left(-1\right)^{n}m\left(x^{n}\right)<b^{-x^{a}}+c$$

All the goal is to determine $\alpha,\beta$

Answer 1

I am going to prove the following inequality :

$\left(\dfrac{1-t}{1+t}\right)^{\!t}>\cos 2t\quad\forall\,t\in(-1,1)\setminus\{0\}\,.\quad\color{blue}{(1)}$

Note that by letting $\;t=\dfrac1x\;$ for any $\;x>1\,,\;$ from $\,(1)\,$ we get the OP’s inequality.

Let $\;f:(-1,1)\to\Bbb R\;$ be the function defined as

$f(t)=(t\!-\!2)\ln(1\!-\!t)-(t\!+\!2)\ln(1\!+\!t)\;\;$ for any $\;t\in(-1,1)\,.$

The function $\,f(t)\,$ is twice differentiable on the interval $\,(-1,1)\,,\,$ moreover $\; f(0)=f’(0)=0\,.$

Since $\;f’’(t)=\dfrac{4t^2}{\left(1-t^2\right)^2}>0\;$ for all $\;t\in(-1,1)\setminus\{0\}\;,\;$ the function $\,f(t)\,$ is convex on $\,(-1,1)\,$.

Consequently, it results that

$f(t)>f(0)+f’(0)\,t=0\;\;$ for any $\;t\in(-1,1)\setminus\{0\}\,.$

Hence ,

$(t\!-\!2)\ln(1\!-\!t)-(t\!+\!2)\ln(1\!+\!t)>0\;$ for any $\,t\in(-1,1)\!\setminus\!\{0\},$

$t\ln\left(\dfrac{1-t}{1+t}\right)>2\ln\left(1-t^2\right)\;\;$ for any $\;t\in(-1,1)\setminus\{0\}\;,$

$\left(\dfrac{1-t}{1+t}\right)^{\!t}>\left(1-t^2\right)^2\;\;$ for any $\;t\in(-1,1)\setminus\{0\}\,.\quad\color{blue}{(2)}$

Analogously, we can prove that the function

$g(t)=\left(1-t^2\right)^2-\cos 2t:(-1,1)\to\Bbb R$

is convex on the interval $\,(-1,1)\,,\,$ moreover $\,g(0)\!=\!g’(0)\!=\!0\,,$

consequently,

$g(t)=\left(1-t^2\right)^2-\cos 2t>0\;\;$ for any $\;t\in(-1,1)\setminus\{0\}\;,$

$\left(1-t^2\right)^2>\cos 2t\;\;$ for any $\;t\in(-1,1)\setminus\{0\}\,.\quad\color{blue}{(3)}$

From the inequalities $\,(2)\,$ and $\,(3)\,,\,$ it follows $\,(1)\,.$

Answer 2

Maybe not quite simpler, but letting $a=1/x$ then $f(a)=\left(\frac{1-a}{1+a}\right)^{a}-\cos\left(2a\right)=\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\cos\left(\frac{2}{x}\right)=g(x)$. Now we consider the Maclaurin series of $f(a)$, more specifically note that:

$$\left(\frac{1-a}{1+a}\right)^{a}>1-2a^2+{4\over3}a^4-{2\over5}a^6$$ and $$1-2a^2+{2\over3}a^4>\cos(2a)$$ so that: $$f(a)>1-2a^2+{4\over3}a^4-{2\over5}a^6-\big(1-2a^2+{2\over3}a^4\big)={94\over3}a^4-288a^6$$ which is positive for $0<a<1$ making $g(x)>0$ for $x>1$.

Answer 3

Same trick as before might give a somewhat shorter solution. Let $x=2/y$, $0<y<\pi/2$. Then $\cos(2/x)=\cos(y)>0$ and by monotonicity of $\ln$ it suffices to show that $$ f(y)=\frac{y}{2}\ln\frac{2/y-1}{2/y+1} - \ln(\cos y)\ge 0. $$ Note that $f(0)=0$ so it will be sufficient to show that $f'(y)\ge 0$. Now, using Taylor series for $\ln(1+a)=a-a^2/2+a^3/3+\dots$, $$ \frac{y}{2}\ln\frac{2/y-1}{2/y+1}= \frac{y}{2}\left[\ln(1-y/2)-\ln(1+y/2)\right]=- \sum_{k=1}^\infty \frac{y^{2k}}{(2k-1)2^{2k-1}} $$ and thus, using the expression for the Taylor series of tangents, \begin{align*} f'(y)&=- \sum_{k=1}^\infty \frac{2k y^{2k-1}}{(2k-1)2^{2k-1}}+\tan(y) \\ & =- \sum_{k=1}^\infty \frac{2k}{(2k-1)2^{2k-1}}y^{2k-1}+ \sum_{k=1}^\infty \frac{2\cdot 4^k(4^k-1)\zeta(2k) }{(2\pi)^{2k}}y^{2k-1} \\ & =\frac{y^3}6 + \frac{23y^5}{240}+\frac{227y^7}{5040}+\dots\ge 0 \end{align*} since for $k\ge 2$ the coefficient corresponding to $y^{2k-1}$ is larger in the second sum: indeed, since $\zeta(2k)=\sum_{i=1}^\infty i^{-2k}>1$ $$ \frac{\frac{2\cdot 4^k(4^k-1)\zeta(2k) }{(2\pi)^{2k}}}{\frac{2k}{(2k-1)2^{2k-1}}}=\left(1-\frac 1{2k}\right)\frac{16^k-4^k}{\pi^{2k}}\zeta(2k)\ge \min_{k=2,3,\dots}\left(1-\frac 1{2k}\right)\left(1-\frac 1{4^k}\right)\left[\frac{4}{\pi}\right]^{2k} =\left(1-\frac 1{4}\right)\left(1-\frac 1{4^2}\right)\left[\frac{4}{\pi}\right]^{4}>1 $$ while for $k=1$ the coefficients are the same.

Answer 4

Just a few remarks since other users gave proofs/

$$f(x)=\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\cos\left(\frac{2}{x}\right)$$ is defined for $x \geq 1$ and does not exist in the real domain otherwise.

Let $x=\frac 1t$ $$g(t)=\left(\frac{1-t}{1+t}\right)^t-\cos (2 t)$$ Let $S_n$ to be the series expansion of $g(t)$ to $O(t^{2n+4})$. The norm $$\Phi_n=\int_0^1\Big[g(t)-S_n\Big]^2\,dt$$ $$\left( \begin{array}{cc} n & \Phi_n \\ 1 & 5.1643\times 10^{-3} \\ 2 & 2.1261\times 10^{-4} \\ 3 & 6.7153\times 10^{-7} \\ 4 & 2.7507\times 10^{-7} \\ 5 & 8.6409\times 10^{-8} \\ \end{array} \right)$$ So, a low order series represent quite well the function. Moreover, we can show that $S_{n}$ does not show any real root.

We could even do much better using, instead of series, the $[4+2n,2n]$ Padé approximant $P_n$ built around $t=0$. Concerning the norms $$\Psi_n=\int_0^1\Big[g(t)-P_n\Big]^2\,dt$$ $$\left( \begin{array}{cc} n & \Psi_n \\ 0 & 5.1643\times 10^{-3} \\ 1 & 2.4784\times 10^{-6} \\ 2 & 9.7386\times 10^{-8} \\ 3 & 2.9313\times 10^{-9} \\ \end{array} \right)$$

In the range of interest, $S_n$ and $P_n$ are $\geq 0$.