An Inequality with Two Variables from Awesome Math
Problem
Solution
By the AM-GM inequality,
$8a^2\sqrt{a}+b^2\sqrt{a}+b^2\sqrt{b}\ge 3\left(8a^2\sqrt{a}\cdot b^2\sqrt{a}\cdot b^2\sqrt{b}\right)^{\frac{1}{3}}=6ab\sqrt{b}.$
The three terms $8a^2\sqrt{a},b^2\sqrt{a},b^2\sqrt{b}$ are never equal as would required for equality. Hence, the inequality is strict.
Illustration
Acknowledgment
I came across this problem at the Lorian Saceanu's Easy Beautiful Math facebook group. The problem was easily traced to the awesome math site's publication Mathematical Reflections (Issue 6 2017, J421.) The problem was proposed by Adrian Andreescu and the solution is by Polyahedra, Polk State College.
The illustration is by Marcos Carreira. Amit Itagi and Nezir Hotic have independently come up with the above solution.
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
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