Minimum value of $\frac{4}{4-x^2}+\frac{9}{9-y^2}$

Question

Given $x,y \in (-2,2)$ and $xy=-1$

Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$

My try:

Converting the function into single variable we get:

$$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$

$$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$

Using Differentiation we get:

$$g'(x)=\frac{8x}{(4-x^2)^2}-\frac{18x}{(9x^2-1)^2}$$

$$g'(x)=2x\left(\frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}\right)$$

$$g'(x)=70x\frac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$

So the critical points are:

$x=0, x=\pm \sqrt{\frac{2}{3}}$

But $x \ne 0$ since $xy=-1$

$$g'(x)=70x \frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$

By using derivative test we get Minimum occurs when $x=\pm \sqrt{\frac{2}{3}}$

Hence $$x^2=\frac{2}{3}, y^2=\frac{3}{2}$$

Min value is $$\frac{4}{4-\frac{2}{3}}+\frac{9}{9-\frac{3}{2}}=\frac{12}{5}$$

Is there any other approach?

Answer 1

Using $xy = -1$, we can put the fractions over a common denominator as $${4\over{4 - x^2}} + {9\over{9 - y^2}} = {{72 - 9x^2 - 4y^2}\over{37 - 9x^2 - 4y^2}} = 1 + {35\over{37 - (9x^2 + 4y^2)}}$$ So we need to minimize $9x^2 + 4y^2$. Since $9x^2 \times 4y^2$ is constant, the minimum occurs when $9x^2 = 4y^2$, i.e. $9x^2 = 4/x^2$, which gives the same result as you found.

Answer 2

With the change of variable $t=x^2$, and differentiating the logarithm of the function, things should be a bit easier:

$$h(t)=\frac4{4-t}+\frac{9t}{9t-1},t\in\left(\frac14,4\right)$$

Then $$h(t)=\frac{-9t^2+72t-4}{(4-t)(9t-1)}$$ $$\log h(t)=\log(-9t^2+72t-4)-\log(4-t)-\log(9t-1)$$ $$(\log h(t))'=\frac{-18t+72}{-9t^2+72t-4}+\frac1{4-t}-\frac9{9t-1}=\frac{315t^2-140}{\text{Irrelevant}}$$ This is $0$ for $t=\frac23$

Answer 3

From “harmonic mean $\le$ arithmetic mean” one gets $$ \frac{2}{f(x, y)} \le \frac12 \left(\frac{4-x^2}{4}+ \frac{9-y^2}{9} \right) = \frac 12 \left(2- \frac{x^2}{4} - \frac{y^2}{9} \right) $$ and from “arithmetic mean $\ge$ geometric mean” $$ \frac{x^2}{4} + \frac{y^2}{9}\ge 2 \frac{|xy|}{6} = \frac 13 $$ so that $$ f(x, y) \ge \frac{4}{2-\frac 13} = \frac{12}{5} \, . $$ In both estimates equality holds if $\frac{x^2}{4} = \frac{y^2}{9} \iff x^2=\frac{2}{3}, y^2=\frac{3}{2},$ i.e. equality is attained the two points $$ ( \sqrt \frac 23, -\sqrt \frac 32) \quad \text{and} \quad ( -\sqrt \frac 23, \sqrt \frac 32) $$ which are both in the given domain $(-2, 2) \times (-2, 2)$ of $f$.

Answer 4

From $xy=-1$

$$ S = \frac{4}{4-x^2} + \frac{9}{9-y^2} = 1 + \frac{35}{25 - \left(3x - \frac{2}{x}\right)^2}\geq \frac{12}{5}$$

Equality hold when $$3x=\frac{2}{x}$$