Minimum value of $\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$

Question

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

Answer 1

$$(a+b)(ax^2+by^2)=p^2+ab$$ 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

Answer 2

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.

Answer 3

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 $S=au$, $T=bv$.

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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.

Answer 4

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*}