While playing with a few two variable inequality and AM-GM inequality, I have ran into the following puzzle:
Question: Show that if $a, b \in (0,1)$ and $a^2+b^2 = 1$, then: $\dfrac{a}{\sqrt{b}} + \dfrac{b}{\sqrt{a}} \ge \sqrt[4]{8}$ with equality when $a = b = \dfrac{1}{\sqrt{2}}$ without using calculus.
I played with it for hours and didn't get to the break point when I can "see" the answer. I used the substitution, AM-GM inequality, and even wolfram alpha to find the min value and I got $\sqrt[4]{8}$. Originally, I planned to only find the min value and after I found it, I set out to prove it without using calculus. So no derivative is allowed. Any cool idea from experts here?
Best Answer
Using the substitutions $a = x^2$ and $b = y^2$ where $x, y \in (0, 1)$, we have $x^4 + y^4 = 1$, so $$ \frac{a}{\sqrt{b}} + \frac{b}{\sqrt{a}} = \frac{x^2}{y} + \frac{y^2}{x} = \frac{x^3 + y^3}{x y}. $$ We would like to show \begin{align*} \frac{x^3 + y^3}{x y} \ge \sqrt[4]{8} &\iff \frac{(x^3 + y^3)^4}{x^4 y^4} \ge 8 (x^4 + y^4) \\ &\iff (x^3 + y^3)^4 \ge 8 (x^8 y^4 + x^4 y^8). \end{align*} Since \begin{multline} (x^3 + y^3)^4 - 8 (x^8 y^4 + x^4 y^8) \\ = (x - y)^2 ( x^{10}+2 x^9 y+3 x^8 y^2+8 x^7 y^3+5 x^6 y^4+2 x^5 y^5 \\ +5 x^4 y^6+8 x^3 y^7+3 x^2 y^8+2 x y^9+y^{10}) \end{multline} where the second factor is positive, the sought inequality holds with equality at $x = y = \frac{1}{2}$.