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Inequality Challenge

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anemone

MHB POTW Director
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Feb 14, 2012
3,690
Prove \(\displaystyle x^x \ge \left( \frac{x+1}{2} \right)^{x+1}\) for $x>0$.
 

Opalg

MHB Oldtimer
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Feb 7, 2012
2,708
Prove \(\displaystyle x^x \ge \left( \frac{x+1}{2} \right)^{x+1}\) for $x>0$.
We want to show that $f(x) = x^x - \bigl( \frac{x+1}{2} \bigr)^{x+1} \geqslant0$ for all $x>0$. Take logs, then differentiate: $$\ln f(x) = x\ln x - (x+1)\ln\bigl( \tfrac{x+1}{2} \bigr),$$ $$\tfrac d{dx}(\ln f(x)) = \ln x + 1 - \ln\bigl( \tfrac{x+1}{2} \bigr) - 1 = \ln\bigl( \tfrac{2x}{x+1} \bigr).$$ But $\ln\bigl( \frac{2x}{x+1} \bigr)$ is zero when $x=1$, negative when $x<1$ and positive when $x>1$. Thus $\ln(f(x))$ has a minimum value when $x=1$, hence so does $f(x)$. But $f(1) = 0$. Therefore $f(x)\geqslant0$ for all $x>0$, as required.

Edit. Oops! Anemone kindly points out a grotesque blunder in the way I presented that argument. Here is what I should have said.
Notice that $x^x \geqslant \bigl( \frac{x+1}{2} \bigr)^{x+1}$ is equivalent to $\dfrac{x^x}{\bigl( \frac{x+1}{2} \bigr)^{x+1}} \geqslant 1$. Take logs to see that this in turn is equivalent to $x\ln x - (x+1)\ln\bigl( \tfrac{x+1}{2} \bigr) \geqslant0$. That can be proved as in my attempt above.
 
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anemone

MHB POTW Director
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Feb 14, 2012
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We want to show that $f(x) = x^x - \bigl( \frac{x+1}{2} \bigr)^{x+1} \geqslant0$ for all $x>0$. Take logs, then differentiate: $$\ln f(x) = x\ln x - (x+1)\ln\bigl( \tfrac{x+1}{2} \bigr),$$ $$\tfrac d{dx}(\ln f(x)) = \ln x + 1 - \ln\bigl( \tfrac{x+1}{2} \bigr) - 1 = \ln\bigl( \tfrac{2x}{x+1} \bigr).$$ But $\ln\bigl( \frac{2x}{x+1} \bigr)$ is zero when $x=1$, negative when $x<1$ and positive when $x>1$. Thus $\ln(f(x))$ has a minimum value when $x=1$, hence so does $f(x)$. But $f(1) = 0$. Therefore $f(x)\geqslant0$ for all $x>0$, as required.

Edit. Oops! Anemone kindly points out a grotesque blunder in the way I presented that argument. Here is what I should have said.
Notice that $x^x \geqslant \bigl( \frac{x+1}{2} \bigr)^{x+1}$ is equivalent to $\dfrac{x^x}{\bigl( \frac{x+1}{2} \bigr)^{x+1}} \geqslant 1$. Take logs to see that this in turn is equivalent to $x\ln x - (x+1)\ln\bigl( \tfrac{x+1}{2} \bigr) \geqslant0$. That can be proved as in my attempt above.
Hi Opalg,

Thanks for participating and I want to show you and those who read this thread another method (a method proposed by other) to prove this inequality using the Jensen Inequality...

Let \(\displaystyle f(x)=x \ln x\)

Differentiate the function of f of x twice we got \(\displaystyle f''(x)=\frac{1}{x} (>0) \) for all $x$.

This means $f(x)$ is a convex function in the domain $x>0$ and Jensen inequality tells us if a function is convex, we have

\(\displaystyle f\left( \frac{\sum x_i}{n}\right) \le \frac{\sum f(x_i)}{n}\)

\(\displaystyle f\left( \frac{x+1}{2}\right) \le \frac{f(x)+f(1)}{2}\)

\(\displaystyle \left( \frac{x+1}{2}\right) \ln \left( \frac{x+1}{2}\right) \le \frac{x\ln x+1 \ln 1}{2}\)

\(\displaystyle \left( \frac{x+1}{2}\right) \ln \left( \frac{x+1}{2}\right) \le \frac{x\ln x}{2}\)

\(\displaystyle \cancel{2}\left( \frac{x+1}{\cancel{2}}\right) \ln \left( \frac{x+1}{2}\right) \le x\ln x\)

\(\displaystyle \ln \left( \frac{x+1}{2}\right)^{x+1} \le \ln x^x\)

\(\displaystyle \therefore x^x \ge \left( \frac{x+1}{2}\right)^{x+1}\)