Minimum value of $\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$ If $$x^2+y^2=1$$
Prove that Minimum value of $$f(x,y)=\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$$ is
$$\frac{2\sqrt{ab}}{a+b}$$
My try:
I used basic Trigonometry:
Let $x=\cos t$ and $y=\sin t$
Then we get $$f(x,y)=g(t)=\frac{a\cos^2 t+b\sin^2 t}{\sqrt{a^2\cos^2 t+b^2\sin^2 t}}$$
Now let $$p=\cos(2t)$$
then we get a single variable function as:
$$h(p)=\frac{1}{\sqrt{2}}\frac{(a+b)+p(a-b)}{\sqrt{a^2+b^2+p(a^2-b^2)}}$$
where $p \in [-1, 1]$
Now we can find critical point and find minimum.
Is there a better approach, i tried lagrange multipliers but very tedious
 A: $$(a+b)(ax^2+by^2)=p^2+ab$$ where $p=\sqrt{a^2x^2+b^2y^2}$
Now assuming $ab>0,$
$$\dfrac{p^2+ab}p\ge2\sqrt{p\cdot\dfrac{ab}p}$$ using AM-GM inequality
A: Since you have
$$x^2 + y^2 = 1 \implies y^2 = 1 - x^2 \tag{1}\label{eq1A}$$
you can substitute this into your $f(x,y)$ to have a function $g(x)$ for $0 \le x \le 1$ with
$$g(x) = \frac{ax^2 + b(1-x^2)}{\sqrt{a^2x^2 + b^2(1-x^2)}} = \frac{(a-b)x^2 + b}{\sqrt{(a^2-b^2)x^2 + b^2}} \tag{2}\label{eq2A}$$
Since $x$ only appears as $x^2$, there's no need to worry about negative values of $x$, so have $0 \le x \le 1$. Assuming $a,b \gt 0$ that $g(0) = g(1) = 1$. Using the quotient rule, you get
$$\begin{equation}\begin{aligned}
g'(x) & = \frac{2(a-b)x\sqrt{(a^2-b^2)x^2 + b^2} - ((a-b)x^2 + b)(2x)(a^2-b^2)\left(\frac{1}{2\sqrt{(a^2-b^2)x^2 + b^2}}\right)}{(a^2-b^2)x^2 + b^2} \\
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
To check for a critical point, setting $g'(x) = 0$ means the numerator must be $0$. Multiplying both sides by $\sqrt{(a^2-b^2)x^2 + b^2}$ and simplifying gives
$$\begin{equation}\begin{aligned}
0 & = 2(a-b)x((a^2-b^2)x^2 + b^2) - ((a-b)x^2 + b)(x)(a^2-b^2) \\
& = x(a-b)(2((a^2-b^2)x^2 + b^2) - ((a-b)x^2 + b)(a + b)) \\
& = x(a-b)(2(a^2 - b^2)x^2 + 2b^2 - (a^2 - b^2)x^2 - (ab + b^2)) \\
& = x(a-b)((a^2 - b^2)x^2 + b^2 - ab)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Thus, you have $x = 0$, $a = b$ or $(a^2 - b^2)x^2 + b^2 - ab$. The only interesting choice is the third one, which I will leave to you to deal with.
A: Let us show
$$
\Big(\ a^2x^2+b^2y^2\ \Big)(x^2+y^2)\cdot\frac{4ab}{(a+b)^2}\le \Big(\ ax^2+by^2\ \Big)^2\ .
$$
It is useful to substitute $u=x^2$, $v=y^2$, so let us show:
$$
4ab\Big(\ a^2u+b^2v\ \Big)(u+v)\le \Big(\ au+bv\ \Big)^2\cdot (a+b)^2 \ .
$$
It turns out that after moving all terms to the R.H.S. we can factor and restate equivalently:
$$
0\le (au-bv)^2(a-b)^2\ ,
$$
which is true. To see this, compare with $(S+T)^2(a+b)^2-(S-T)^2(a-b)^2=(\dots)^2-(\dots)^2=4(Sa+Tb)(Ta+Sb)$, where $S=au$, $T=bv$.

Note: The inequality of Cauchy-Schwarz gives:
$$
\Big(\ (ax)^2+(by)^2\ \Big)(x^2+y^2)\ge \Big(\ ax^2+by^2\ \Big)^2\ .
$$
so the maximum of the given expression is one.
A: Note that
\begin{eqnarray*}
f(x,y) = \frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}  \\
\end{eqnarray*}
\begin{eqnarray*}
= \frac{ \sqrt{ab}}{a+b} \left(  \sqrt{ \frac{a^2x^2+b^2y^2}{ab}}  + \sqrt{ \frac{ab}{a^2x^2+b^2y^2}} \right) 
\end{eqnarray*}
\begin{eqnarray*}= \frac{ \sqrt{ab}}{a+b} \left(  \left( \sqrt[4]{ \frac{a^2x^2+b^2y^2}{ab}}  - \sqrt[4]{ \frac{ab}{a^2x^2+b^2y^2}} \right)^2+2 \right). 
\end{eqnarray*}
