Quadratika - Inspired by Sandeep Sir!

Let $f(y) = y^2 - my +1$

$f(1)<0 \Rightarrow m>2$

Let $z= \frac{4|x|}{x^2 + 16} = \frac{4t}{t^2 + 16} ....(t=|x|)$

$\Rightarrow zt^2 -4t +16z =0$

Since ' $t$ ' is real, $D \geq 0$

$\Rightarrow 0 \leq z \leq 0.5$ as $z \geq 0$

$0 \leq z^m \leq \frac{1}{2^m} \Rightarrow [z^m]=0$

Firstly m>2 .

Next as we are working in positive domain remove mod sign .

Next denominator is x+16/x which has a minimum value of 16 by AM-GM.

So The thing whose power is greater than 2 is less than 1 . hence greatest integer will give 0.

As simple as that!

$f(y) = y^2 - my + 1$

Since the given parabola is opening upward and 1 lies between the roots of the equation $f(y) = 0$ ,

$f(1) < 0 \Rightarrow 1 - m + 1 < 0 \Rightarrow \boxed{m>2}$

The integral has non-negative limits, hence we could use A.M - G.M .

$\dfrac{x^2 + 16}{2} \geq \sqrt{x^2(16)} \Rightarrow \dfrac{4|x|}{x^2 +16} \leq \dfrac{1}{2}$

$\lfloor (\frac{4|x|}{x^2 + 16})^m \rfloor = 0 . . . . . . . . . \because a^m < 1 \text{ when } a < 1 \text{ \& } m > 0$

So the integrand is zero.

Rest is trivial.