A Trigoquality

Let $x\sin x=a$ .

$\frac{9x^2sin ^{ 2 }{ x }+4}{x\sin x}=\frac{9a^2+4}{a}$

We wish to minimize this, and there are a number of ways to do this. Personally, I prefer calculus. Look for critical point:

$\frac{d}{dx}(\frac{9a^2+4}{a})=0$

$9-\frac{4}{a^2}=0$

$a=\frac{2}{3}$

You find that the minimum occurs at $a=\frac{2}{3}$ , and we know it's possible for $x\sin x = \frac{2}{3}$ . Plugging that back into the equation gets us $\boxed{12}$ .

We can rewrite the expression $\frac{4+9x^{2}\sin{x}^{2}}{x \sin{x}}$ as: $\frac{4}{x \sin{x}}+9x\sin{x}$ Since $x>0$ , we can find the minimum by using AM-GM: $\frac{\frac{4}{x \sin{x}}+9x\sin{x}}{2}\geq\sqrt{36}$ $\frac{4}{x \sin{x}}+9x\sin{x}\geq12$ So we find the minimum: $\boxed{12}$ .

taking 6 outside it becomes 6((3xsinx)/2+2/(3xsinx)) it is of the form 6(y+1/y) where y=(3/2) xsinx min value of y+1/y=2 [AM>=GM] this gives min value =6 2=12 also y=1 => xsinx=2/3 which is possible under the given limits and xsinx is not equal to zero within the given limits

This is a problem from the 1983 AIME. Check the solutions out here.

Let us define $h$ as $x\sin x$ .

We can then set this equal to its lowest value, which we'll call $k$ , so we'd have $\frac{4 + 9h^2}{h} = k$ .

Rearranging everything to the left hand side of the equation gives us $9h^2 - kh + 4 = 0$

In the quadratic formula, the discriminant is under a square root, so the minimum value of the discriminant must be $0$ .

Setting the discriminant equal to $0$ gives us $k^2 - (4)(9)(4) = 0 \Rightarrow k^2 = 144$ .

So we have either $k = 12$ or $k = -12$ . We know that $x \sin x$ must be positive between $0$ and $\pi$ , since both $x$ and $\sin x$ are positive in this interval, so we have $\boxed{k = 12}$ .

Max value of sinx is 1 So the given exp becomes as (9x^2+4)/x Now it can be simplified as 3*2((3x/2)+(2/3x)) = 3x2x2 =12