An Inequality with Exponents
Problem
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
- An Inequality with Just Two Variable $\left(\displaystyle\left(\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\right)\left(\frac{a+b}{2ab} + \sqrt{\frac{2}{a^2+b^2}}\right) \le \frac{(a+b)^2}{ab}\right)$
- An Inequality with Just Two Variable II $\left(\displaystyle\small{\left(\frac{2ab}{a+b}+\sqrt{ab}+\frac{a+b}{2}\right)\left(\frac{a+b}{2ab} + \frac{1}{\sqrt{ab}}+\frac{2}{a+b}\right) \le 5 +2 \left(\frac{a}{b}+\frac{b}{a}\right)}\right).$
- An Inequality with Just Two Variable III $\left(\displaystyle\frac{a}{b\sqrt{2}}+\frac{b\sqrt{2}}{a}+2\left(\frac{\sqrt{a^2+b^2}}{b}+\frac{b}{a^2+b^2}\right)\ge \frac{9\sqrt{2}}{2}\right)$
- An Inequality with Just Two Variables IV $\left(\displaystyle\frac{a+2}{a^2+a+1}+\frac{b+2}{b^2+b+1}\ge \frac{ab+2}{(ab)^2+ab+1}+1\right)$
- An Inequality with Just Two Variables V $\left(\displaystyle \left(\ln\left(\frac{1}{2a}+\frac{1}{2b}\right)\right)^{a+b}\ge \left(\ln\frac{1}{a}\right)^b\left(\ln\frac{1}{b}\right)^a\right)$
- An Inequality with Just Two Variables VI $\left(|x-y|(x+1)(y+1)\le 2\right)$
- An Inequality with Just Two Variable VII $\left(\displaystyle (x^3+y^3)^3(x^2-xy+y^2)\ge x^2y^2\sqrt{xy}(x^2+y^2)^3\right)$
- An Inequality with Just Two Variable VIII $\left(\displaystyle \frac{a^2}{b+2}+\frac{b^3}{a+2}+(2-a)b^2\le 12\right)$
- The power of substitution III: proving an inequality with two variables $\left(\displaystyle\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\le\frac{2}{\sqrt{1+ab}}\right)$
- Simple Yet Uncommon Inequalities with Absolute Value $\left((|x|-|y|)^2\le |x^2-y^2|,\,|\sqrt{|x|}-\sqrt{|y|}|\le\sqrt{|x-y|}\right)$
- An Inequality with Just Two Variable And an Integer $\small{\left(\displaystyle \Bigr(\frac{a}{b^n}+\frac{b}{a^n}\Bigr)\Bigr(\frac{a^n}{b}+\frac{b^n}{a}\Bigr)\Bigr(\frac{a^n}{b^n}+\frac{b}{a}\Bigr)\Bigr(\frac{b^n}{a^n}+\frac{a}{b}\Bigr)\geq 8\left(\sqrt{\displaystyle\left(\frac{a}{b}\right)^{n-1}}+\sqrt{\displaystyle\left(\frac{b}{a}\right)^{n-1}}\right)\right)}$
- An Inequality with Exponents $\left(\displaystyle b^b\cdot e^{a+\frac{1}{a}}\ge 2e^b\right)$
- Problem 790 from Pentagon: an Inequality in Two Variables $\left(\displaystyle\ln \left|\left(\frac{2+\sin 2b}{2+\sin 2a}\right)\right|\leq \frac{2\sqrt{3}}{3}(b-a)\right)$
- An Inequality with Two Variables from Awesome Math $\left(\displaystyle \frac{6ab-b^2}{8a^2+b^2}\lt\sqrt{\frac{a}{b}}\right)$
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