Inequality : $\Big(\frac{x^n+1+(\frac{x+1}{2})^n}{x^{n-1}+1+(\frac{x+1}{2})^{n-1}}\Big)^n+\Big(\frac{x+1}{2}\Big)^n\leq x^n+1$

Question

I have the following problem to solve :

Let $x,y>0$ and $n>1$ a natural number then we have : $$\Big(\frac{x^n+y^n+(\frac{x+y}{2})^n}{x^{n-1}+y^{n-1}+(\frac{x+y}{2})^{n-1}}\Big)^n+\Big(\frac{x+y}{2}\Big)^n\leq x^n+y^n$$

The problem is equivalent to : $$\Big(\frac{x^n+1+(\frac{x+1}{2})^n}{x^{n-1}+1+(\frac{x+1}{2})^{n-1}}\Big)^n+\Big(\frac{x+1}{2}\Big)^n\leq x^n+1$$ Or ($y^2=x$): $$\Big(\frac{y^{2n}+1+(\frac{y^2+1}{2})^n}{y^{2(n-1)}+1+(\frac{y^2+1}{2})^{n-1}}\Big)^n+\Big(\frac{y^2+1}{2}\Big)^n\leq y^{2n}+1$$ I try the following identity : $$ch^2(x)-sh^2(x)=1$$

So we put $y=sh(x)$ we get : $$\Big(\frac{sh^{2n}(x)+1+(\frac{ch^2(x)}{2})^n}{sh^{2(n-1)}(x)+1+(\frac{ch^2(x)}{2})^{n-1}}\Big)^n+\Big(\frac{ch^2(x)}{2}\Big)^n\leq sh^{2n}(x)+1$$

And after I'm stuck...

Update case $n=3$ :

Due to homogeneity we can assume : $$x^{2}+y^{2}+\Big(\frac{x+y}{2}\Big)^{2}=1$$

Remains to show :

$$\Big(x^3+y^3+\Big(\frac{x+y}{2}\Big)^3\Big)^3+\Big(\frac{x+y}{2}\Big)^{3}-x^3-y^3\leq 0$$

Or :

$$\frac{1}{512} (x + y) (729 x^8 + 972 x^6 y^2 + 1728 x^5 y^3 + 54 x^4 y^4 + 1728 x^3 y^5 + 972 x^2 y^6 - 448 x^2 + 640 x y + 729 y^8 - 448 y^2)\quad(1)$$

Or :

$$\frac{1}{512} (x + y)(27 (x + y)^2 (3 x^2 - 2 x y + 3 y^2)^3-448y^2-448z^2+640xy)$$

Or :

$$\frac{1}{512} (x + y)(27 (x + y)^2 (3 x^2 - 2 x y + 3 y^2)^3-64 (7 x^2 - 10 x y + 7 y^2))$$

But with the constraint :

$$\frac{5x^2}{4}+\frac{xy}{2}+\frac{5y^2}{4}=1$$

Or :

$$x^2+y^2=\Big(1-\frac{xy}{2}\Big)\frac{4}{5}$$

It gives :

$$\frac{1}{512} (x + y)\Big(27 (x + y)^2 \Big(\Big(1-\frac{xy}{2}\Big)\frac{12}{5}-2xy\Big)^3-64 \Big(\Big(1-\frac{xy}{2}\Big)\frac{28}{5}-10xy\Big)\Big )$$

Or :

$$\frac{1}{512} (x + y)\Big(27 \Big(\Big(1-\frac{xy}{2}\Big)\Big)\frac{4}{5}+2xy) \Big(\Big(1-\frac{xy}{2}\Big)\frac{12}{5}-2xy\Big)^3-64 \Big(\Big(1-\frac{xy}{2}\Big)\frac{28}{5}-10xy\Big)\Big)$$

We put the substitution $a=xy$ .There is a root at $a=\frac{1}{3}$ it gives :

$$\frac{1}{512} (x + y)\Big(-\frac{512}{625} (3 a - 1) (576 a^3 - 816 a^2 + 52 a - 73)\Big)$$

Now with the constraint it's not hard to see that $a\leq \frac{1}{3}$

And $$f(a)=(576 a^3 - 816 a^2 + 52 a - 73)\leq 0$$ on $[0,\frac{1}{3}]$

So the quantity $(1)$ is negative . We are done for this case .

If you have a hint it would be cool .

Thanks a lot !

Answer 1

Let's begin with your 2nd inequality with $x$ only.

Let $x = (1+p)/(1-p)$ and we have $x>0\iff p\in(-1,1)$, then your inequality becomes

$$\left(\frac{(1+p)^n+(1-p)^n+1}{(1+p)^{n-1}+(1-p)^{n-1}+1}\right)^n+1\le\!\!\!?\;(1+p)^n+(1-p)^n.$$

Let $a_n = (1+p)^n+(1-p)^n$, by symmetry we can assume $p\in(0,1)$ (the case $p=0$ is trivial), then we have

$$\left(\frac{a_n+1}{a_{n-1}+1}\right)^n+1\le\!\!\!?\;a_n.$$

Note that $\left(\frac{a_n+1}{a_{n-1}+1}\right)$ is increasing [1], we have

$$\left(\frac{a_n+1}{a_{n-1}+1}\right)^n\le\left(\frac{a_n+1}{a_{n-1}+1}\right)\cdot\left(\frac{a_{n+1}+1}{a_{n}+1}\right)\cdots\left(\frac{a_{2n-1}+1}{a_{2n-2}+1}\right)=\left(\frac{a_{2n-1}+1}{a_{n-1}+1}\right).$$

Thus, your inequality can be proved if we have

$$\frac{a_{2n-1}+1}{a_{n-1}+1}\le\!\!\!?\;a_n-1,$$

Note that $(a_n-1)(a_{n-1}+1)=a_{2n-1}+2(1-p^2)^{n-1}+a_n-a_{n-1}-1$, so the inequality above becomes

$$(1-p^2)^{n}+\frac12(a_{n+1}-a_n)\ge\!\!\!?\;1.$$

Let $(u_n)$ be the LHS, we have $u_0=u_1=1$, and we can prove that $(u_n)$ is increasing [2], which finishes the proof.

PS. In fact, a stronger inequality has been proved. The following equalities can be useful: $$a_{n+1}=a_n+pb_n,\quad b_{n+1}=b_n+pa_n,$$ with $b_n=(1+p)^n-(1-p)^n$.

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Update.

[1] \begin{align} \frac{a_{n+1}+1}{a_n+1}\ge\frac{a_{n}+1}{a_{n-1}+1} & \iff(a_{n+1}+1)(a_{n-1}+1)\ge(a_n+1)^2\\ &\impliedby a_{2n}+a_2(1-p^2)^{n-1}+1\ge a_{2n}+2(1-p^2)^n+1\tag{*}\\ &\iff (1+p)^2+(1-p)^2 \ge 2(1+p)(1-p)\\ &\iff (2p)^2\ge0. \end{align}

(*): we have $a_{n+1}+a_{n-1}\ge2a_n$ since

$$a_{n+1}+a_{n-1}\ge2a_n\iff b_{n}\ge b_{n-1}\iff a_{n-1}\ge 0.$$

[2] \begin{align} u_{n+1}\ge u_{n} & \iff \frac p2 b_{n+1}+(1-p^2)^{n+1}\ge\frac p2 b_{n}+(1-p^2)^{n}\\ & \iff\frac {p^2}2a_{n}\ge p^2(1-p^2)^{n}\\ & \iff \frac1{(1-p)^n}+\frac1{(1+p)^n}\ge2.\\ \end{align}

As mentioned by @Andreas, the last inequality is true since $$\frac1{(1-p)^n}+\frac1{(1+p)^n}\ge \frac2{(1-p^2)^{n/2}}\ge2.$$

Answer 2

Consider the inequality in the form

$$\Big(\frac{x^n+y^n+(\frac{x+y}{2})^n}{x^{n-1}+y^{n-1}+(\frac{x+y}{2})^{n-1}}\Big)^n+\Big(\frac{x+y}{2}\Big)^n\leq x^n+y^n $$ Due to homegeneity, let $\frac{x+y}{2} = 1$. Denote $\frac{x-y}{2} = a$. Then we consider the equivalent question, w.l.o.g. with $0\le a \le 1$: $$ \Big(\frac{(1+a)^n+(1-a)^n+1}{(1+a)^{n-1} +(1-a)^{n-1} + 1 }\Big)^n \leq- 1+ (1+a)^n+(1-a)^n $$

Note that for every convex function $f(x)$ and $p_i \ge 0$ the following is true (Jensen):

$$ f\Big( \frac{p_1 x_1 + p_2 x_2 + p_3 x_3 }{p_1 + p_2 + p_3 }\Big) \le \frac{p_1 f(x_1) + p_2 f(x_2) + p_3 f(x_3) }{p_1 + p_2 + p_3 } $$

Since $f(x) = x^n$ is a convex function, apply Jensen to the LHS with $p_1 = (1+a)^{n-1}$, $p_2 = (1-a)^{n-1}$, $p_3 = 1$, and $x_1 = 1+a$, $x_2 = 1-a$, $x_3 = 1$. Then it is sufficient to prove

$$ \frac{(1+a)^{n-1} (1+a)^{n} + (1-a)^{n-1} (1-a)^{n} + 1}{(1+a)^{n-1} +(1-a)^{n-1} + 1 } \le - 1+ (1+a)^n+(1-a)^n $$ Clearing denominators gives $$ g(a,n) = -(2 + (1+a)^n+(1-a)^n) + (1+a)^n(1+(1-a)^{n-1}) + (1-a)^n(1+(1+a)^{n-1}) \;\\ \; = 2 (1 - a^2)^{n-1} + a ((1+a)^{n-1} - (1-a)^{n-1} ) -2 \ge 0 $$ But $g(a,n)$ is, for every $n \ge 2$, a function which is monotonously increasing in $a$. Namely, $g(a=0,n) = 0$ and $g(a=1,n) = 2^{n-1} -2$.

This can be proved by induction over $n$. We have that the assumption holds for $n=2$ since $g(a,n=2) = 0$ $\forall a$. Do the step $n \to n+1$. We need to show $$ g(a, n+1) = 2 (1 - a^2)^{n} + a ((1+a)^{n} - (1-a)^{n} ) -2 \ge 0 $$ This can be rewritten $$ g(a, n+1) = g(a, n) + a^2 [(1+a)^{n-2} + (1-a)^{n-2} - 2((1+a)(1-a))^{n-2}] \ge 0 $$

Since $g(a, n) \ge 0$ by induction hypothesis, this holds true if the second bracket is nonnegative. Use the AM-GM inequality for the first two terms in this bracket, giving the stronger condition $2 ((1+a)(1-a))^{(n-2)/2} - 2((1+a)(1-a))^{n-2} \ge 0 $ or $(1-a^2)^{(n-2)/2} \le 1 $ which is obviously true.

This proves the claim. $\qquad \Box$

Remark: This inequality is rather fine-tuned. Consider as a first factor $$\Big(\frac{x^n+y^n+z\cdot(\frac{x+y}{2})^n}{x^{n-1}+y^{n-1}+z\cdot(\frac{x+y}{2})^{n-1}}\Big)^n $$ and let $z$ increase from 0 to 1. It is easy to see that the increase of $z$ makes the term smaller. Choosing $z=1$ makes the term "just small enough" for the inequality to be "$\le$". Indeed, for $z=0$, this inverses to "$\ge$", as this post (with this proof) shows.

Numerically, one can solve for the particular value $z^*$ where equality occurs. It shows that for all $n$, we have $\lim_{a \to 0} z^* = 0$. The highest value for $z^*$ occurs for $n=2$ and $a=1$ with $z^* = \sqrt 3 - 1 \simeq 0.732$. For $n \ge3$ and all $a$, we have that $z^* < 0.423$.