Find minimum of $a+b$ under the condition $\frac{m^2}{a^2}+\frac{n^2}{b^2}=1$ where $m,n$ are fixed arguments

Question

Assume $m,n \in \mathbb{R}$ is fixed. And $a,b(a>b>0)$ satisfied the equation $$\frac{m^2}{a^2}+\frac{n^2}{b^2}=1$$ Find $\min\{a+b\}$

Answer 1

Let $a=m\sec t$ and $b=n\csc t$ where $t\in (0,\frac{\pi}{2})$

\begin{align*} f(t) &= m\sec t+n\csc t \\ f'(t) &=m\sec t \tan t-n\csc t \cot t \\ \end{align*}

$$f'(t)=0 \implies \frac{n}{m}=\tan^{3} t$$

\begin{align*} \min(a+b) &= \left( \frac{m}{\sqrt[3]{m}}+\frac{n}{\sqrt[3]{n}} \right) \sqrt{m^{2/3}+n^{2/3}} \\ &= \left( m^{2/3}+n^{2/3} \right)^{3/2} \end{align*}

The answer let me thinking of the longest ladder problem.

Further point to be noticed:

At $\displaystyle t=\tan^{-1} \sqrt[3]{\frac{n}{m}}$,

$$\frac{a}{b} = \frac{m}{n} \sqrt[3]{\frac{n}{m}}$$ $$\frac{a^{3}}{b^{3}}=\frac{m^{2}}{n^{2}} $$ $$a>b>0 \implies m^{2}>n^{2}$$ When $m^{2}<n^{2}$ and $0<b<a$, $a+b$ will be bounded below by $2\sqrt{m^2+n^2}$ which corresponds to $m\sec t=n\csc t$.

Answer 2

Wlog

$a=m\sec t,b=n\csc t$

$(a+b)^2\ge4ab=\dfrac{4mn}{\sin t\cos t}$

The equality occurs if $a=b$

i.e., if $m\sec t=n\csc t$

$\frac m{\cos t}=\frac n{\sin t}=\pm\sqrt{m^2+n^2}$

Answer 3

Hint:
By Holder's inequality, $$\left(\frac{m^2}{a^2}+\frac{n^2}{b^2} \right)(a+b)^2 \geqslant (m^{2/3}+n^{2/3})^3$$