how to prove $(x+1)^{(x-1)} \leq x^x$?

Question

I need to prove the statement:

$$\forall x \ge 1, \quad \log_{2}(x!) - \frac{1}{2}x\log_2(x) \geq 0$$

I tried with a proof by induction (if there is an easier way, let me know!), so I tested the basis case $x = 1$, which holds, and by induction hypothesis, assumed that $\log_{2}(x!) - \frac{1}{2}x\log_2(x) \geq 0$ holds. Now I want to come to the conclusion that $$\log_{2}((x+1)!) - \frac{1}{2}(x+1)\log_2(x+1) \geq 0$$

After some manipulations, I find that $$\log_{2}((x+1)!) - \frac{1}{2}(x+1)\log_2(x+1) = \log_{2}(x!) - \frac{1}{2}\left[x\log_2(x+1)-\log_2(x+1)\right]$$ So if $$x\log_2(x+1)-\log_2(x+1) \leq x\log_2(x)$$ then $$\log_{2}((x+1)!) - \frac{1}{2}(x+1)\log_2(x+1) \geq 0$$

Again, after some manipulations, I come to this inequality $(x+1)^{(x-1)} \leq x^x$. I know it is true, but I can't figure out how to prove it.

score 2 · Accepted Answer · answered Apr 27 '20 at 17:38

2

Substitute $x \to x + 1$. Then, $(x+2)^{x} \leq (x+1)^{x}\cdot (x+1)\implies (1+\frac{1}{x+1})^{x} \leq x + 1$. This can be proven by induction.

The $x=1$ case is trivial, as $\frac{3}{2}\leq 2$.

Assume $(1+\frac{1}{k+1})^k \leq k+1$ is true. Then

$(1+\frac{1}{k+2})^{k+1} < (1+\frac{1}{k+1})^{k+1}\leq(k+1)(1+\frac{1}{k+1})=k+1+1=k+2$

Therefore by mathematical induction, the statement $(x+1)^{x-1}\leq x^x$ is true.

answered Apr 27 '20 at 17:38

Gareth Ma

1,881

Induction won't work (at least by itself), because we need to prove it for all $x \geqslant 1$. – Botond Apr 27 '20 at 17:49
In the question the equation to be proven is from $\log_2(x!)\cdots$, so I assumed it's only needed for natural numbers. I'll keep this here anyways. – Gareth Ma Apr 27 '20 at 17:51
You are right, I missed that part. – Botond Apr 27 '20 at 17:52
Yes, I forgot to specify, $x \in \mathbb{N}^+$. – Philip Gaudreau Apr 27 '20 at 18:05

score 2 · Answer 2 · answered Apr 27 '20 at 17:41

2

Your equation is equivalent to $$\left(1+\frac{1}{x}\right)^{x-1}\le x,$$ which is true for $x=1$ (since $1\le1$), for $x=2$ (since $\frac{3}{2}\le2$), and true for higher $x$ since $$\left(1+\frac{1}{x}\right)^{x-1}<e<x.$$ $\blacksquare$

answered Apr 27 '20 at 17:41

ViHdzP

4,854
2
20
51

Quanto · Answer 3 · 2020-04-27T19:18:39.110

1

For $x\ge 1$, examine

\begin{align} f(x) =& x\ln x-(x-1)\ln(1+x) \\ = & f(1)+\int_1^x f’(t)dt \\ = & \>0+\int_1^x \left(f’(1)+\int_1^t f’’(s)ds\right)dt\\ =&\int_1^x \left( 1-\ln2+\int_1^t \frac{1-s}{s(1+s)^2}ds \right) dt\ge 0 \\ \end{align}

where $f’(1)=1-\ln2>0$ and $f’’(x)= \frac{1-x}{x(1+x)^2}\ge0 $. Thus,

$$x\ln x\ge (x-1)\ln(1+x) \implies(x+1)^{(x-1)} \leq x^x$$

edited Apr 27 '20 at 19:18

answered Apr 27 '20 at 18:46

Quanto

120,125

score 0 · Answer 4 · answered Apr 28 '20 at 04:41

Without induction.

Using natural logarithms, you want to prove that $$\Delta=2 \log (x!)-x \log (x) \geq 0$$

Using Stirling approximation $$\Delta=x (\log (x)-2)+\log (2 \pi x)+\frac{1}{6x}+O\left(\frac{1}{x^3}\right)$$ which is positive as soon as $x \geq 2$.

how to prove $(x+1)^{(x-1)} \leq x^x$?

4 Answers4