Prove that ${\left(\frac{x+1}{2}\right)}^{x+1} \leq x^x$ I want to prove that for each real number $x>0$, the statement
$${\left(\frac{x+1 } {2}\right)}^{\!x+1} \leq x^x$$
is correct. 
I’ve tried using mathematical induction methods, but haven’t found a solution. Although, using any method is appreciated.
 A: I'll try to prove this by induction itself.
For $n=1$;
$$P(1):={\left(\frac{1+1 } {2}\right)}^{\!1+1} \leq 1^1$$
$$1^2 \leq 1^1$$
This holds true. Let it be true for $P(k)$ as well.
$$P(k):={\left(\frac{k+1 } {2}\right)}^{\!k+1} \leq k^k\tag{Ineq. 01}\label{1}$$
And for $P(k+1)$;
$$P(k+1):={\left(\frac{k+2 } {2}\right)}^{\!k+2} \leq (k+1)^{k+1}\tag{Ineq. 02}\label{2}$$
By Ineq. $01$; we know that;
$$(k+1)^{k+1} \leq 2^{k+1}k^k$$
Substituting that with Ineq. $02$;
$$(k+2)^{k+2} \leq 2^{k+2}2^{k+1}k^k$$
$$(k+2)^{k+2} \leq 8(4k)^k$$
$$\frac{(k+2)^{k+2}}{(4k)^k} \leq 8$$
$$\left(\frac{1}{4}+\frac{1}{2k}\right)^k(k+2)^2 \leq 8$$
$$\left(\frac{1}{4}+\frac{1}{2k}\right)^k \leq \frac{8}{k^2}+2(1+\frac{2}{k})$$
We know that $\left(\frac{1}{4}+\frac{1}{2k}\right)^k<1$
Hence, this is true. And The claim is true for all $n ∈ N$
A: You can show your inequality using the concavity of the logarithm:


*

*$(\star)$: For $p \in [0,1]$ and $x,y>0$ you have 
$$p\log x + (1-p) \log y \leq \log (px+(1-p)y)$$
Hence,
$${\left(\frac{x+1 } {2}\right)}^{\!x+1} \le x^x \Leftrightarrow \left(\frac{x+1}{x}\right)^x(x+1)\leq 2^{x+1}$$
$$\Leftrightarrow x\log \left(\frac{x+1}{x}\right) + \log (x+1)\leq (x+1)\log 2$$
$$\stackrel{(\star)}{\Leftrightarrow} \frac{x}{x+1}\log \left(\frac{x+1}{x}\right) + \frac{1}{x+1}\log (x+1)$$ $$\stackrel{(\star)}{\leq} \log \left(\frac{x}{x+1}\cdot \frac{x+1}{x} + \frac{1}{x+1}\cdot (x+1) \right) = \log 2$$
The last inequality is true because of $(\star)$.
A: We need to prove that
$$x\ln{x}\geq(x+1)\ln\frac{x+1}{2}$$ or
$$\frac{x\ln{x}+1\ln1}{2}\geq\frac{x+1}{2}\ln\frac{x+1}{2},$$ which is true by Jensen for a convex function $f(x)=x\ln{x}.$
A: you can use functions for that: lets assume $$f(x)= x^x - (\frac{x+1}{2})^{x+1}$$. then, its derviative (after simplification) is: $$f'(x)= −(x+1)^{x+1}⋅2^{−x−1}(\ln(x+1)+1)+x^x(\ln(x)+1)+\ln(2)(x+1)^{x+1}⋅2^{−x−1}$$.
you can check that $f(0)=\frac{1}{2}$, $f(1)=0$, and the value of $f'(x)>0$ for all $x>1$, and $f'(x)<0$ for all $0<x<1$.
there for the function is always equal or above the x-axis and therefore the inequality stands
