<s>[INST] 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
 [/INST] 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.
</s>