# Prove that $\left(\frac{x^n+1}{x^{n-1}+1}\right)^n+\left(\frac{x+1}{2}\right)^n\geq x^n+1$

Let $$x>0$$ and $$n$$ be a natural number. Prove that: $$\left(\frac{x^n+1}{x^{n-1}+1}\right)^n+\left(\frac{x+1}{2}\right)^n\geq x^n+1.$$

This question is very similar to many contests problems, but I think it's much more harder than contest problem and it's just impossible to solve this problem during a competition.

In my fifth comment I wrote:

but I think there is no a chance that it will be solved there.

I tried to use $$M_n^2\geq M_{n+1}M_{n-1},$$ where $$x>0$$ and $$M_n=\sqrt[n]{\frac{x^n+1}{2}}$$, but without success.

I think a perspective way to solve this problem it's something like the following wrong way:

https://mathoverflow.net/revisions/337457/1

Thank you!

• General claim with $a,b,c,d$ is certainly false already for $n=2$: say, check $(5,4,7,1)$. – Fedor Petrov Aug 2 at 6:00
• BTW I am curious where this problem came up. It seems rather finely tuned. – GH from MO Aug 2 at 11:45
• Over at math.se this was tagged "contest math". If this actually is from a contest, then it should not be posed as a problem here until after the contest is over. Does anyone know anything more about this? – Todd Trimble Aug 2 at 22:51
• @Yemon Choi My name is Michael. I don't think that this problem from a contest. Just this problem is very similar to many contest problems. I think this problem is very hard for a contest. – Michael Rozenberg Aug 4 at 18:37
• Is it ok to post on MO a question another user posted on MSE without even mentioning in the question that it was somebody else's idea? – Pierre-Yves Gaillard Aug 7 at 12:22

Corrected proof, see GH from MO's comment and answer: A generalization of the inequality gives more flexibility for variations of parameters, which eventually yields a proof. One observation is $$\frac{x+1}{2}=\frac{x^b+1}{x^{b-1}+1}$$ for $$b=1$$. If one wants to achieve $$(\frac{x^a+1}{x^{a-1}+1})^n+(\frac{x^b+1}{x^{b-1}+1})^n\geq x^n+1$$ for all $$x\ge0$$, looking at derivatives at $$x=1$$ shows that a necessary condition is $$a+b-1\ge n$$. In fact, the following generalization holds true for all real $$a,b\ge1$$ and $$x\ge0$$: $$\begin{equation} \left(\frac{x^a+1}{x^{a-1}+1}\right)^{a+b-1}+\left(\frac{x^b+1}{x^{b-1}+1}\right)^{a+b-1}\geq x^{a+b-1}+1. \end{equation}$$ Setting $$n=a+b-1$$, $$a=(n+1)/2+c/2$$, $$x=y^2$$ and $$z=y^c$$, this inequality is equivalent to $$\begin{equation} \left(\frac{y^{n+1}+z}{y^{n-1}+z}\right)^n+\left(\frac{y^{n+1}+1/z}{y^{n-1}+1/z}\right)^n\geq y^{2n}+1, \end{equation}$$ where $$y>0$$ is arbitrary and $$z>0$$ is in an interval depending on $$y$$ and $$n$$. However, the inequality holds true for all $$y>0$$ and $$z>0$$. By the $$z\leftrightarrow\frac{1}{z}$$ and $$y\leftrightarrow\frac{1}{y}$$ symmetries, we assume in the following $$0.

Fix $$y$$ and $$n$$. We show that the left hand side is monotonically increasing in $$z$$. Taking that for granted the assertion follows, for we have equality for $$z=0$$.

Taking the derivative with respect to $$z$$, the monotonicity is equivalent to $$\begin{equation} \left(\frac{y^{n+1}+z}{zy^{n+1}+1}\right)^{n-1}\ge \left(\frac{y^{n-1}+z}{zy^{n-1}+1}\right)^{n+1}. \end{equation}$$ (In a previous version, there was a miscalculation, observed by GH from MO, which made the rest of the "proof" easier.)

Raising both sides to the $$\frac{1}{(n-1)(n+1)}$$-th power, and fixing now $$y$$ and $$z$$, the claim follows once we know that $$\begin{equation} (0,\infty)\to\mathbb R,\;\;t\mapsto\left(\frac{y^t+z}{zy^t+1}\right)^{\frac{1}{t}} \end{equation}$$ is monotonically increasing in $$t$$. For this GH from MO's answer contains an elegant proof. Here is another one:

Take the derivative (with respect to $$t$$) of the logarithm of this function, then multiply by $$t^2$$ and set $$w=y^t$$. The result is $$\begin{equation} h(w,z):=\log w\cdot w\cdot(\frac{1}{w+z}-\frac{z}{zw+1})+\log(zw+1)-\log(w+z). \end{equation}$$ We need to show that $$h(w,z)\ge0$$ for all $$w,z\in(0,1)$$. The derivative of $$h(w,z)$$ with respect to $$w$$ is $$\begin{equation} \frac{\partial h(w,z)}{\partial w}=\frac{(1-w^2)(1-z^2)z\log w}{((zw+1)(w+z))^2}<0. \end{equation}$$ So $$h(w,z)$$, for fixed $$z$$, is decreasing in $$w$$. Thus $$h(w,z)\ge h(1,z)=0$$.

• On the right hand side of your third display, a factor of $\frac{1}{z^2}$ is missing. That is, monotonicity is equivalent to $\left(\frac{y^{n+1}+z}{zy^{n+1}+1}\right)^{n-1}\ge \left(\frac{y^{n-1}+z}{zy^{n-1}+1}\right)^{n+1}$. I added below a detailed proof of this inequality, for any $n\in[1,\infty)$ and $y,z\in[0,1]$. – GH from MO Aug 10 at 4:18
• $\frac{x+1}{2}=\frac{x^b+1}{x^{b-1}+1}$ it's $\left(x^{b-1}-1\right)(x-1)=0,$ which for $b\neq1$ gives $x=1$ only. – Michael Rozenberg Aug 10 at 10:17
• @GHfromMO Thank you for spotting the miscalculation and fixing the proof! I added another argument for the final monotonicity assertion. – Peter Mueller Aug 10 at 11:15
• @MichaelRozenberg: Peter Mueller proved the inequality contained in his first display. Setting $a:=n$ and $b:=1$ in this inequality, you obtain the inequality in your post. Alternatively, start from his second display, and set $y:=x^{1/2}$ and $z:=x^{(n-1)/2}$ there. – GH from MO Aug 10 at 11:49
• It's an extremely nice proof! Thank you Peter! @GH from MO Thank you! – Michael Rozenberg Aug 10 at 18:28

This is a supplement (correction) to Peter Mueller's nice solution. As he observed, it suffices to show that, for any fixed $$n\geq 1$$ and $$y\in[0,1]$$, the function $$z\mapsto\left(\frac{y^{n+1}+z}{y^{n-1}+z}\right)^n+\left(\frac{zy^{n+1}+1}{zy^{n-1}+1}\right)^n,\qquad z\in(0,1),$$ is increasing. (Indeed, $$y:=x^{1/2}$$ and $$z:=x^{(n-1)/2}$$ yields the LHS of the OP's inequality, while $$y:=x^{1/2}$$ and $$z:=0$$ yields the RHS of the OP's inequality.) Taking the derivative with respect to $$z$$, the statement becomes $$\left(\frac{y^{n+1}+z}{zy^{n+1}+1}\right)^{n-1}\ge \left(\frac{y^{n-1}+z}{zy^{n-1}+1}\right)^{n+1},\qquad y,z\in(0,1).$$ Let us now fix $$y,z\in(0,1)$$ and think of $$n\geq 1$$ as the variable. Taking the logarithm of both sides and dividing by $$(n-1)(n+1)$$, it suffices to show that the function $$t\mapsto\frac{1}{t}\log\frac{y^t+z}{zy^t+1},\qquad t>0,$$ is increasing. Making the change of variable $$w:=y^t$$, it suffices to show that the function $$w\mapsto\frac{\log(w+z)-\log(wz+1)}{\log w},\qquad w\in(0,1),$$ is increasing. Writing $$w=:\tanh u$$ and $$z=:\tanh v$$, it suffices to show that the function $$u\mapsto\frac{\log\tanh(u+v)}{\log\tanh(u)},\qquad u>0,$$ is increasing. Taking the derivative with respect to $$u$$, the statement becomes $$\sinh(u)\cdot\cosh(u)\cdot\log\tanh(u)\geq\sinh(u+v)\cdot\cosh(u+v)\cdot\log\tanh(u+v).$$ That is, it suffices to show that the function $$u\mapsto \sinh(u)\cdot\cosh(u)\cdot\log\tanh(u),\qquad u>0,$$ is decreasing. With the notation $$s:=-\log\tanh(u)$$, we have $$\sinh(u)\cdot\cosh(u)\cdot\log\tanh(u)=\frac{e^{-s}}{\sqrt{1-e^{-2s}}}\cdot\frac{1}{\sqrt{1-e^{-2s}}}\cdot(-s)=\frac{-s}{2\sinh s},$$ hence it suffices to show that the function $$s\mapsto\frac{\sinh s}{s},\qquad s>0$$ is increasing. However, this is clear, because the Taylor series of this function converges everywhere, and it has nonnegative coefficients.

$$\newcommand{\s}{\overset{\text{sgn}}=} \newcommand{\Dx}{\text{Dx}} \newcommand{\logDx}{\text{logDx}} \newcommand{\DlogDx}{\text{DlogDx}} \newcommand{\DDDlogDx}{\text{DDDlogDx}} \newcommand{\DDDDDlogDx}{\text{DDDDDlogDx}} \newcommand{\dif}{\text{dif}} \newcommand{\Ddif}{\text{Ddif}} \newcommand{\R}{\mathbb{R}}$$ Let us show that the inequality in question holds for all real $$n\ge5$$; the cases when $$n\in\{1,2,3,4\}$$ are verified directly. By a comment of Pietro Mayer, without loss of generality $$0. We shall reduce the problem to the completely algorithmic problem of checking sign patterns of several polynomials in $$n,x$$, of total degrees $$\le11$$. This reduction is done in a few steps:

Step 1: Eliminating $$(\frac{1+x}2)^n$$: The inequality in question can be rewritten as $$\begin{equation} u(x):=u_n(x):=n \ln \left(\frac{x^n+1}{x^{n-1}+1}\right) -\ln \left(x^n+1-z^n\right)\ge0, \end{equation}$$ where $$z:=z_x:=\frac{1+x}2$$. Note that $$\begin{multline*} u'(x)\frac{x (1+x)}n \left(x^{1-n}+x^n+x+1\right) \left(x^n+1-z^n\right) \\ =\Dx:=\left(n \left(1-x^2\right)+\left(x^{2-n}-1\right) \left(1+x^n\right)\right) z^n-(n-1) \left(1-x^2\right) \left(1+x^n\right), \end{multline*}$$ so that $$\begin{equation} u'(x)\s\Dx\s\logDx(x), \end{equation}$$ where $$\s$$ denotes the equality in sign and
$$\begin{equation} \logDx(x):=\logDx_n(x):=n \ln z-\ln \frac{(n-1) \left(1-x^2\right) \left(1+x^n\right)}{n \left(1-x^2\right)+\left(x^{2-n}-1\right) \left(1+x^n\right)}. \end{equation}$$ Here and in the sequel, $$\Dx$$, $$\logDx$$, etc. are atomic, "indivisible" symbols; $$\Dx$$ refers to the derivative (of $$u$$) in $$x$$, $$\logDx$$ refers to a certain kind of logarithmic modification of $$\Dx$$, etc. Next, let $$\begin{multline*} \DlogDx(x):=\DlogDx_n(x):= \\ \logDx'(x)(1-x) (1+x) x^{n-1} \left(1+x^n\right) \left(n \left(1-x^2\right)+\left(x^{2-n}-1\right) \left(1+x^n\right)\right) \\ =n^2 (x-1)^2 (x+1) \left(x-x^n\right) x^{n-2}-2 \left(x^n-1\right) \left(x^n+1\right)^2+\frac{n (x-1) \left(x^n+1\right)^2 \left(x^n+x\right)}{x}. \end{multline*}$$ So, we get a polynomial in $$x^n$$ of degree $$3$$ over the field $$\R(n,x)$$ of all real rational functions in $$n,x$$.

Step 2: Reducing the degree from $$3$$ to $$2$$: Let $$\begin{multline*} \DDDlogDx(x):= \DlogDx''(x) x^{3 - 3 n}\\ =x^{3-3 n} (n (n x-n+2 x+2) (n^2 x^2-n^2+n x^2+2 n-1) x^{n-3} \\ -2 (n-1) n (x-1) (2 n^2 x^2-2 n^2+n x^2-2 n x+3 n+2 x) x^{2 n-4}+n (3 n-1) (3 n x-3 n-6 x+2) x^{3 n-3}) \\ \s\DlogDx''(x). \end{multline*}$$ Taking the second derivative $$\DlogDx''(x)$$ of the polynomial $$\DlogDx(x)$$ in $$x^n$$ of over $$\R(n,x)$$ kills the free term of that polynomial. Thus, we get the polynomial $$\DDDlogDx(x)$$ of degree $$2$$ in $$x^{-n}$$ of over $$\R(n,x)$$.

Step 3: Reducing the degree from $$2$$ to $$1$$: Let $$\begin{equation} \DDDDDlogDx(x):= \frac{\DDDlogDx''(x)}{2 (n - 1) n^2 x^{-3 - 2 n}} = A_n(x) - x^n B_n(x), \end{equation}$$ $$\begin{equation} A_n(x):=\left(2 n^3+3 n^2-5 n-6\right) x^4+\left(-2 n^3+3 n^2+3 n-2\right) x^3+\left(-2 n^3-n^2+5 n-2\right) x^2+\left(2 n^3-5 n^2+n+2\right) x, \end{equation}$$ $$\begin{equation} B_n(x):=2 n^3+3 n^2-\left(-2 n^3+5 n^2-n-2\right) x^3-\left(2 n^3+n^2-5 n+2\right) x^2-\left(2 n^3-3 n^2-3 n+2\right) x-5 n-6, \end{equation}$$ so that $$\begin{equation} \DDDlogDx''(x)\s A_n(x) - x^n B_n(x). \end{equation}$$ Thus, we get the polynomial $$\DDDDDlogDx(x)$$ of degree $$1$$ in $$x^n$$ of over $$\R(n,x)$$.

Step 4: Reducing the degree from $$1$$ to $$0$$:
We can see that (under the conditions $$n\ge5$$ and $$0, assumed everywhere here) $$B_n(x)>0$$. So, $$\DDDlogDx''(x)<0$$ whenever $$A_n(x)\le0$$.

Further, let $$\begin{equation} \dif(x) = \dif_n(x) :=\ln\frac{A_n(x)}{B_n(x)} - n \ln x\s A_n(x) - x^n B_n(x)\s \DDDlogDx''(x) \end{equation}$$ wherever $$A_n(x)>0$$, and then $$\begin{multline*} \Ddif(x) = \Ddif_n(x) :=\dif'(x)\frac{A_n(x)B_n(x)}{(n+1)(n-2)} \\ =-4 n^5 (x-1)^4 (x+1)^2+4 n^4 (x-1)^4 (x+1)^2+n^3 (x-1)^2 \left(15 x^4+16 x^3-10 x^2+16 x+15\right) \\ -4 n^2 \left(x^2-1\right)^2 \left(5 x^2+x+5\right)+n \left(-x^6+30 x^5+41 x^4-44 x^3+41 x^2+30 x-1\right) \\ +2 \left(3 x^6-6 x^5-11 x^4-36 x^3-11 x^2-6 x+3\right) \\ \s\dif'(x), \end{multline*}$$ finally getting a polynomial in $$n,x$$.

Now we need to trace the above steps back:

Looking back at the polynomial $$A_n(x)$$, (for $$x\in(0,1)$$) we find that $$A_n(x)\le0$$ iff $$x_1\le x\le x_2$$, where $$x_1=x_1(n)$$ and $$x_2=x_2(n)$$ are the two roots of $$A_n(x)$$ in $$(0,1)$$ such that $$x_1.

Further, $$\Ddif<0$$ and hence $$\dif'<0$$ on $$(0,x_1]$$; and $$\Ddif>0$$ and hence $$\dif'>0$$ on $$[x_2,1)$$. So, $$\dif$$ decreases on $$(0, x_1]$$ and increases on $$[x_2, 1)$$. So, $$\dif$$ is $$+-$$ on $$(0, x_1]$$ (that is, $$\dif$$ can switch sign at most once on $$(0, x_1]$$, and only from $$+$$ to $$-$$). Similarly, $$\dif$$ is $$-+$$ on $$[x_2, 1)$$.

But also $$\dif(1)=0$$. So, actually $$\dif<0$$ on $$[x_2, 1)$$.

So, $$\DDDlogDx''$$ is $$+-$$ on $$(0, x_1]$$ and $$\DDDlogDx'' < 0$$ on $$[x_2, 1)$$.

Also, $$A < 0$$ and hence $$\DDDlogDx'' < 0$$ on $$[x_1, x_2]$$. So, $$\DDDlogDx''$$ is $$+-$$ on $$(0, 1)$$. So, $$\DDDlogDx$$ is convex-concave on $$(0, 1)$$. Also, $$\DDDlogDx(1)=0$$.

So, $$\DDDlogDx$$ is $$+-+$$ on $$(0, 1)$$. So, $$\DlogDx$$ is convex-concave-convex on $$(0, 1)$$. Also, $$\DlogDx(1)=\DlogDx'(1)=\DlogDx''(1)=0>-8n(n^2-1)=\DlogDx'''(1)$$ and $$\DlogDx(0+)=2-n<0$$. So, $$\DlogDx$$ is $$-+$$; so, $$\logDx$$ is decreasing-increasing.

Also, $$\logDx(1-)=0$$. So, $$\logDx$$ is $$+-$$, and hence so is $$\Dx$$ (with $$z = \frac{1 + x}2$$).

Recalling that $$u'(x)\s\Dx$$, we see that $$u_n(x)$$ is increasing-decreasing (in $$x\in(0,1)$$). Also, $$u_n(0)=-\ln(1 - 2^{-n})>0$$ and $$u_n(1)=0$$.

Thus, $$u>0$$, which concludes the proof.

• I have highlighted the steps of the reduction to the polynomial, which should make the idea clearer. – Iosif Pinelis Aug 7 at 17:32

Mathematica confirms that for concrete values of $$n$$, e.g.

n = 30; ForAll[x,x > 0, ((x^n+1)/(x^(n-1)+1))^n + ((x + 1)/2)^n >= x^n + 1];Resolve[%,Reals]


True

Mathematica fails with it in the general case of an arbitrary positive integer $$n$$.

• Mathematica confirms such cases as $n=\pi$ too. – user64494 Aug 5 at 5:38
• I looked for a human's proof. – Michael Rozenberg Aug 25 at 4:45