Inequality $\frac{a}{\sqrt{b}} + \frac{b}{\sqrt{a}} \ge \sqrt[4]{8}$ on the unit circle

Question

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?

Answer 1

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 $x = y = \frac{1}{2}$.

Answer 2

Let $a+b=2u\sqrt{ab}$.

Thus, by AM-GM $u\geq1$ and we need to prove that: $$\left(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}\right)^4\geq8(a^2+b^2)$$ or $$(\sqrt{a}+\sqrt{b})^4(a+b-\sqrt{ab})^4\geq8a^2b^2(a^2+b^2)$$ or $$(2u+2)^2(2u-1)^4\geq8(4u^2-2),$$ which is true by AM-GM: $$(2u+2)^2(2u-1)^4=(2u+2)^2\cdot(2u-1)^2\cdot(2u-1)^2\geq16\cdot1\cdot(2u^2-1)=8(4u^2-2).$$ Something stronger.

Let $a$ and $b$ be positive numbers such that: $a^9+b^9=2$. Prove that: $$\frac{a^2}{b}+\frac{b^2}{a}\geq2.$$

Generalization:

let $a$ and $b$ be positives and $n$ be a natural such that $a^{(2n+1)^2}+b^{(2n+1)^2}=2$. Prove that: $$\frac{a^{n+1}}{b^n}+\frac{b^{n+1}}{a^n}\geq2.$$