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- Feb 14, 2012
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Prove \(\displaystyle x^x \ge \left( \frac{x+1}{2} \right)^{x+1}\) for $x>0$.
Prove \(\displaystyle x^x \ge \left( \frac{x+1}{2} \right)^{x+1}\) for $x>0$.
Hi Opalg,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.