An Inequality with Exponents

Problem

An Inequality with Exponents

Solution 1

Part 1

Observe that

$\displaystyle\begin{align} b\ln b+a +\frac{1}{a}-\ln 2-b&=b\ln b+2+\left(a+\frac{1}{a}-2\right)-\ln 2-b\\ &\ge \small{b\ln b+\left(a+\frac{1}{a}-2\right)+2+1-e^{\ln 2}-b}&\text{(*)}\\ &\ge b(\ln b-1)+\left(a+\frac{1}{a}-2\right)+1&\text{(**)}\\ &\ge b\left(\frac{b-1}{b}-1\right)+\left(a+\frac{1}{a}-2\right)+1\\ &=a+\frac{1}{a}-2\ge 0.&\text{(***)}. \end{align}$

  • (*) since $e^{\ln 2}\ge 1+\ln 2,$
  • (**) since $\displaystyle \ln (1+x)\ge \frac{x}{1+x},$ for $x\ge 0,$
  • (***) since $0\lt b\le 1$ and $\displaystyle a+\frac{1}{a}\ge 2.$

Hence, $b\ln b+a+\frac{1}{a}\ge \ln 2+b,$ implying $b^b\cdot a^{a+\frac{1}{a}}\ge 2e^b.$

Part 2

Observe that

$\displaystyle\begin{align} b\ln b+a +\frac{1}{a}-b\ln 2-b&=b\ln b+2+\left(a+\frac{1}{a}-2\right)-b\ln 2-b\\ &\ge \small{b\ln b+2+\left(a+\frac{1}{a}-2\right)+b(1-e^{\ln 2})-b}&\text{(*)}\\ &\ge \small{b\left(\frac{b-1}{b}\right)+2(1-b)+\left(a+\frac{1}{a}-2\right)}&\text{(**)}\\ &= \small{1-b+\left(a+\frac{1}{a}-2\right)\ge 0.}&\text{(***)}. \end{align}$

  • (*) since $e^{\ln 2}\ge 1+\ln 2,$
  • (**) since $\displaystyle \ln (1+x)\ge \frac{x}{1+x},$ for $x\ge 0,$
  • (***) since $0\lt b\le 1$ and $\displaystyle a+\frac{1}{a}\ge 2.$

Hence, $b\ln b+a+\frac{1}{a}\ge b\ln 2+b,$ implying $b^b\cdot a^{a+\frac{1}{a}}\ge (2e)^b.$

Solution 2

Part 1

The minimum for $\displaystyle a+\frac{1}{a}$ is 2, for $a=1$, so we need to prove $b^b e^2\ge 2 e^b$.

Since $\displaystyle \log(x) \geq \frac{x-1}{x}$, we need to prove $2-log(2)\geq 1$, which is satisfied since $\log(x)\geq x-1$ for $x$ positive.

Part 2

As before, $e^2 b^b\geq 2^b e^b$. Since $\displaystyle x^x\geq \frac{1}{2-x}$ for $x \in (0,1]$,

$\displaystyle \frac{e^2}{2-b}\geq 2^b e^b.$

Taking logs

$2- \log(2-b) \geq b (1+\log (2))$

Since $\log(x) \leq x-1$,

$1\geq b \log(2)$

which is true since $b\leq 1$.

Acknowledgment

Dan Sitaru has kindly posted at the CutTheKnotMath facebook page an inequality by Abdallah El Farissi. The inequality was previously published at the Romanian Mathematical Magazine. Solution is by Diego Alvariz; Solution 2 is by N. N. Taleb.

 

Inequalities in Two Variables

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