When $x$ is a real number and $x>1$, why is $x^x>(x+1)^{x-1}$?

When $x$ is a real number and $x>1$, why is the following true?

$x^x>(x+1)^{x-1}$

I tried finding the minimum of $x^x-(x+1)^{x-1}$ with my limited calculus knowledge, but it shortly appeared out of my range.

It's good when I can understand a good answer, but I'd still be happy to come back years later when I'm better at math, so please don't hesitate to share your knowledge.

Let $f(x)=x\ln x-(x-1)\ln(x+1),\;$ so $\color{red}{f(1)=0}$

and $\displaystyle f^{\prime}(x)=x\left(\frac{1}{x}\right)+\ln x-(x-1)\cdot\frac{1}{x+1}-\ln(x+1)=\frac{2}{x+1}-(\ln(x+1)-\ln x)$.

Since $\displaystyle \ln(x+1)-\ln x<\frac{1}{x}\;\;$ (by considering the area under $y=\frac{1}{x}$ from $x$ to $x+1$),

$\displaystyle \color{red}{f^{\prime}(x)>\frac{2}{x+1}-\frac{1}{x}=\frac{x-1}{x(x+1)}>0}\;$ for $x>1$;

so for $x>1,\;\;$ $\color{red}{f(x)>0}\implies x\ln x>(x-1)\ln(x+1)\implies x^x>(x+1)^{x-1}$

By applying the weighted AM-GM to the distinct positive numbers $x+1$ and $1$ with weights $1-\tfrac1x$ and $\tfrac1x$, $$(x+1)^{1-\frac1x}\cdot 1^{\frac1x} < \big(1-\tfrac1x\big)\cdot (x+1) + \tfrac1x\cdot 1 = x.$$ Taking $x$th powers, $$(x+1)^{x-1} < x^x.$$

• Nothing beats this answer +1... – Paramanand Singh Jan 18 '16 at 11:38
• How did you see this? O_O – user85798 Jan 24 '16 at 2:45
• Sekots: it is very similar to $(1+\tfrac1{n+1})^{n+1}>(1+\frac1n)^n$. – user141614 Jan 27 '16 at 14:36

use that the inequality is equivalent to $x+1>\left(1+\frac{1}{x}\right)^x$

• I don't see why the latter is true… – Bernard Jan 15 '16 at 18:41
• think about the Eulerian number – Dr. Sonnhard Graubner Jan 15 '16 at 18:51
• I can see it is asymptotically true, but I don't see why it is true forall $x>1$. – Bernard Jan 15 '16 at 19:26

Observe that if we consider $f(x)=x^x-(x+1)^{x-1}=e^{x\ln x}-e^{(x-1)\ln (x+1)}$

Then $f'(x)>0$ and $f(1)=0$.

So we can conclude that $f(x)>0$.

Hope this helps.

• $f(0) is not defined. How do you know$f'(x)>0$? – Bernard Jan 15 '16 at 18:34 • Actually,$f(x)<0$for$0<x<1$; and$f(1)=0$, but since the OP specified that the inequality is valid for$x>1$, the derivative argument applies (with these corrections). – Fede Poncio Jan 15 '16 at 19:57 The generalized version of Bernoulli's Inequality says that for$y\gt x\gt0\begin{align} 1+y &=1+\tfrac yxx\\ &\lt(1+x)^{\frac yx} \end{align} Therefore, $$(1+y)^{\frac1y}\lt(1+x)^{\frac1x}$$ That is(1+x)^{\frac1x}$is a strictly decreasing function. Therefore, for$x\gt1$, $$(1+x)^{\frac1x}\lt(1+(x-1))^{\frac1{x-1}}$$ and $$(1+x)^{x-1}\lt x^x$$ Let$x>1$. Use the inequalities $$1-\frac1x < \log(x) < x-1.$$ Then $$\log(x+1)=\log(x)+\log(1+\frac1x)<\log(x)+\frac1x.$$ Now multiply by the positive number$x-1\$: $$(x-1)\log(x+1)<x \log(x) + 1-\frac1x - \log(x) < x \log(x).$$