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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.

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

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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, $$ (x+1)^{x-1} < x^x. $$

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

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

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  • $\begingroup$ $f(0) is not defined. How do you know $f'(x)>0$? $\endgroup$ – Bernard Jan 15 '16 at 18:34
  • $\begingroup$ 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). $\endgroup$ – Fede Poncio Jan 15 '16 at 19:57
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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 $$ (1+x)^{x-1}\lt x^x $$

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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 $x-1$: $$(x-1)\log(x+1)<x \log(x) + 1-\frac1x - \log(x) < x \log(x).$$

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