# Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$

Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $$x \ge 1$$ and for all positive integers $$m$$ and $$k \le 2m+1$$.

My work. If $$k \le 2m+1$$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}.$$ I need prove that for all real $$x \ge 1$$ and for all positive integers $$m$$ the inequality $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$$ holds.

• What have you tried? Induction? Jun 7 '19 at 19:11
• Doesn't this also hold for $m = 0$? Jun 7 '19 at 19:12
• Failed to solve the problem by induction. Jun 7 '19 at 19:14

Since by PM $$\left(\frac{x^n+1}{2}\right)^k\geq\left(\frac{x^k+1}{2}\right)^n$$ is true for all $$x\geq0$$ and $$n\geq k>0,$$ it's enough to prove that $$\frac{x^{m+1}+1}{x^m+1}\geq\sqrt[2m+1]{\frac{x^{2m+1}+1}{2}}$$ or $$f(x)\geq0,$$ where $$f(x)=\ln\left(x^{m+1}+1\right)-\ln\left(x^m+1\right)-\frac{1}{2m+1}\ln\left(x^{2m+1}+1\right)+\frac{\ln2}{2m+1}.$$ Indeed, $$f'(x)=\frac{(m+1)x^m}{x^{m+1}+1}-\frac{mx^{m-1}}{x^m+1}-\frac{x^{2m}}{x^{2m+1}+1}=$$ $$=\frac{x^{m-1}\left(mx^{2m+2}-(m+1)x^{2m+1}+(m+1)x-m\right)}{\left(x^m+1\right)\left(x^{m+1}+1\right)\left(x^{2m+1}+1\right)}.$$ Let $$g(x)=mx^{2m+2}-(m+1)x^{2m+1}+(m+1)x-m.$$

Thus, $$g'(x)=m(2m+2)x^{2m+1}-(m+1)(2m+1)x^{2m}+m+1;$$ $$g''(x)=2m(m+1)(2m+1)x^{2m}-2m(m+1)(2m+1)x^{2m-1}=$$ $$=2m(m+1)(2m+1)x^{2m-1}(x-1)\geq0,$$ which says $$g'(x)\geq g'(1)=0,$$ which says $$g(x)\geq g(1)=0,$$ which gives $$f(x)\geq f(1)=0$$ and we are done!

By the way, we see that your inequality is true for all reals $$x\geq0$$ and $$m>0$$.

• Very nice proof! Thank you very much! Jun 11 '19 at 17:23
• @Witold You are welcome! Jun 11 '19 at 17:24
• I was waiting for you, Michael. Jun 11 '19 at 17:43
• @Piquito I just thought that it was proven before. By the way, see the Geethu Joseph's solution. It's something interesting. Jun 11 '19 at 17:45
• I think this does not affect the proof. Sorry Jun 12 '19 at 0:04

The result can be proved using AM $$\geq$$ GM.

\begin{align} \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}&=\sqrt[2m+1]{\prod_{i=0}^{2m} \frac{1+x^{i+1}}{1+x^{i}}}\leq\frac{1}{2m+1} \sum_{i=0}^{2m} \frac{1+x^{i+1}}{1+x^{i}} \\ &=\frac{1}{2m+1} \left[\sum_{i=0}^{m-1} \left(\frac{1+x^{i+1}}{1+x^{i}}+\frac{1+x^{2m-i+1}}{1+x^{2m-i}} \right)+\frac{1+x^{m+1}}{1+x^m}\right]\\ &=\frac{1}{2m+1}\left[ \sum_{i=0}^{m-1} \left( 2x- \frac{x-1}{1+x^{i}} -\frac{x-1}{1+x^{2m-i}}\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\ &=\frac{1}{2m+1}\left[ \sum_{i=0}^{m-1}\left( 2x- \frac{x-1}{1+x^m}\left[\frac{1+x^m}{1+x^{i}}+\frac{1+x^m}{1+x^{2m-i}}\right]\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\ &=\frac{1}{2m+1}\left[ \sum_{i=0}^{m-1}\left( 2x- \frac{x-1}{1+x^m}\left[2+\frac{x^i(x^m-1)(x^{m-i}-1)^2}{(1+x^i)(1+x^{2m-i})}\right]\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\ &\leq \frac{1}{2m+1}\left[ \sum_{i=0}^{m-1}\left(2x-(x-1)\frac{2}{1+x^{m}}\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\ &=\frac{1+x^{m+1}}{1+x^{m}}. \end{align}

• Check please the step from fourth to fifth line. Jun 11 '19 at 17:52
• @Witold There is more mistake, but now it's not so relevant. Jun 11 '19 at 18:24
• Thanks, both! Corrected the mistake. Jun 12 '19 at 5:18
• Very nice proof! Thank you very much! P.S. Here (a) follows because $$\frac{1}{1+x^i}+\frac{1}{1+x^{2m-i}} \ge \frac{2}{1+x^m} \Leftrightarrow \\ (1+x^m)(2+ x^{2m-i}+ x^i) - 2(1+x^i)( 1+x^{2m-i}) \ge 0$$ and $(1+x^m)(2+ x^{2m-i}+ x^i) - 2(1+x^i)( 1+x^{2m-i})=x^i(x^m-1)(x^{m-i}-1)^2$. Jun 12 '19 at 19:28

Let $$f(x,k)=\frac{x^{m+1}+1}{x^m+1}-\left(\frac{x^k+1}2\right)^{\frac1k}\tag1$$ so that $$f_x(x,k)=\frac{\partial f(x,k)}{\partial x}=\frac{x^{m-1}(x^{m+1}+(m+1)x-m)}{(x^m+1)^2}-\frac{x^{k-1}}{x^k+1}\left(\frac{x^k+1}2\right)^{\frac1k}\tag2.$$ Clearly, $$f_x(1,k)=\frac2{2^2}-\frac12=0\tag3$$ and since \begin{align}\small\!\!\!\!\!\!\!\!\!\!\!\!f_{xx}(x,k)=\frac{\partial^2f(x,k)}{\partial x^2}&=\small\frac{mx^{m-2}(x^m+1)^2(2x^{m+1}+m(m+1)x-m(m-1))}{(x^m+1)^4}\\&\small \,\,\,\,\,\,\,\,-\frac{2mx^{2m-2}(x^{m+1}+(m+1)x-m)}{(x^m+1)^4}-\frac{(k-1)x^{k-2}}{(x^k+1)^2}\left(\frac{x^k+1}2\right)^{\frac1k}\tag4,\end{align} we have that $$f_{xx}(1,k)=\frac{m\cdot2^2(2+2m)-2m\cdot2}{2^4}-\frac{k-1}{2^2}=\frac{m(1+2m)-k+1}{2^2}>0\tag5$$ for $$m\ge1$$ due to the fact that $$k<2m+2$$. Therefore, $$x=1$$ is a minimum with $$f(1,k)=\frac22-1=0\tag6.$$ It remains to show that $$f(x,k)>0$$ for $$x>1$$. However, it suffices to prove that $$f_m(x)=\frac{x^{m+1}+1}{x^m+1}-\left(\frac{x^{2m+1}+1}2\right)^{\frac1{2m+1}}>0\tag7$$ since $$\left(\frac{x^k+1}2\right)^{\frac1k}\le\left(\frac{x^{2m+1}+1}2\right)^{\frac1{2m+1}}\tag8$$ from your attempt. Now $$(7)$$ is implied by $$2(x^{m+1}+1)^{2m+1}>(x^m+1)^{2m+1}(x^{2m+1}+1)\tag9$$ which is equivalent to $$2\sum_{n=0}^{2m+1}\binom{2m+1}nx^{nm+n}>(x^{2m+1}+1)\sum_{n=0}^{2m+1}\binom{2m+1}nx^{nm}\tag{10}.$$ Collecting the summation terms yields $$\sum_{n=0}^{2m+1}\left[\binom{2m+1}nx^{nm}(2x^n-x^{2m+1}-1)\right]>0\tag{11}$$ which is almost certainly the case.

By the power mean inequality, with $$x\geq 0, n \geq k$$, we have

$$\left( \frac{ x^n + 1 } {2} \right) ^\frac{1}{n} \geq \left( \frac{x^k + 1 } { 2} \right) ^ \frac{1}{k}.$$

Hence, all that we need to show is:

$$\frac{ x^{m+1} + 1 } { x^m + 1} \geq \left( \frac{ x^{2m+1 } + 1 } {x^0 + 1 } \right) ^ \frac{1}{2m+1} \quad (1) .$$

We will show that

$$\left(\frac{1+x^{m+1}}{1+x^m}\right)^2 \geq \frac{1+x^{m+1+i}}{1+x^{m+i}} \times \frac{1+x^{m+1-i}}{1+x^{m-i}} \quad (2)$$

By cross multiplying and expanding terms (Thanks Wolfram), this is equivalent to:

$$x^{m-i} (x^i-1)^2(x-1)(x^{2m+1} - 1) \geq 0.$$

This is obviously true for $$x \geq 0$$.
(Easier to see by splitting up into cases $$x\geq 1, x < 1$$, which might explain where that condition came from.)

Then, (1) follows by multiplying the chain of inequalities of (2), going from $$i=0$$ to $$m$$.

Note: The (pseudo-)reasoning behind why (2) should be true, is that when we write it as $$\prod ( 1+ x^{a_i}) \geq \prod ( 1+ x^{b_i})$$, with $$a_i, b_i$$ in decreasing order, we have $$\sum a_i = \sum b_i$$. Since $$a_1 = m+i < m+i+1 = b_1$$, it follows that after cancelling the $$x^{\sum a_i}$$ term on both sides, the LHS has a larger exponent, so it will be true for large enough $$x$$.

• You write: "This inequality follows by proving the chain of inequalies, for each k=2m+1 (which gives the LHS) to 1 (which gives the RHS)". Can you show exactly how this inequality follows? Jun 11 '19 at 21:03
• I understood it. I didn’t understand what you wanted to say when you said that "This inequality follows by proving the chain of inequalies, for each k=2m+1 (which gives the LHS) to 1 (which gives the RHS)" Jun 11 '19 at 21:09
• Why inequality $$( x^0 + 1 ) ( x^{m+1} + 1)^{2m+1} \geq (x^{2m+1} + 1) (x^{m} + 1)^{2m+1}$$ follows from inequality $$(x^{2m+1 - k} + 1)(x^m+1)^{2m+1-k}(x^{m+1}+1)^k \geq (x^{2m+1-k-1}+1)(x^m+1)^{2m+1-k-1} ( x^{m+1}+1)^{k+1}$$ ? Jun 11 '19 at 21:17
• OK. If $m=1,k=2, x=2$ then $x^m+x^{2m+1-k} \geq x^{m+1} +x^{2m+1 - k - 1 } \Leftrightarrow 2+2 \ge 4+1$ Jun 11 '19 at 21:30
• @Witold Alright, here's a new version. It was modified from my old version after accounting for the wrong half of the inequality chain. It is essentially Geethu's solution, though it slightly demystifies the AM-GM step. Jun 12 '19 at 1:50

for all $$m\in \mathbb{N}$$ and $$x\geq 1$$ we have $$x^{2m+1}\geq 1$$ then $$2x^{2m+1}\geq x^{2m+1}+1$$

Notice that $$\frac{x^{m+1}+1}{x^m+1} = x - \frac{x-1}{x^m+1} \geq x$$ because $$x\geq 1$$

Then $$2\left(\frac{x^{m+1}+1}{x^m+1}\right)^{2m+1}\geq 2x^{2m+1}\geq x^{2m+1}+1$$

We obtain easily that $$\frac{x^{m+1}+1}{x^m+1} \geq \sqrt[2m+1]{\frac{x^{2m+1}+1}{2}}$$

• Mistake. $\frac{x^{m+1}+1}{x^m+1} \le x$ because $x \ge 1$ Jun 7 '19 at 19:41