How to prove if $5/2 < x < (5/4)(1+\sqrt2)$, then $25/(x(2x-5)\ge 8$

Question

if $\frac52 < x < \frac54(1+\sqrt2)$, then $\frac{25}{x(2x-5)} \ge 8$

First I unpacked the conclusion to:

$$ 16w^2-40w-25 \le 0 $$

I attempted to solve by manipulating the interval (squaring, multiplying):

$$100 \lt 16w^2 ≤25(3+2\sqrt2) $$

$$-50(1+\sqrt2)≤-40w <-100 $$

Then I added the inequalities and subtracted 25.

$$ 50-50\sqrt2-25<16w^2-40w-25<-25+50\sqrt2-25 $$

but that give me

$$ 16w^2-40w-25 \lt 50(\sqrt2-1) $$ which is not $ 16w^2-40w-25 \le 0 $

what did I do wrong?

(there is another interval that i already proved)

Answer 1

$\frac52 < x < \frac54(1+\sqrt2), then \frac{25}{x(2x-5)} \ge 8 $

The conclusion is the same as $x(2x-5) \le 25/8 $ (since $x > 5/2$)

or $4x^2-10x \le 25/4 $

or $(2x-5/2)^2-25/4 \le 25/4 $

or $(2x-5/2)^2-25/4 \le 25/4 $

or $(2x-5/2)^2 \le 25/2 $

or $|2x-5/2| \le 5/\sqrt{2} =5\sqrt{2}/2 $

or $-5\sqrt{2}/2 \le 2x-5/2 \le 5\sqrt{2}/2 $

or $5/2-5\sqrt{2}/2 \le 2x \le 5/2+5\sqrt{2}/2 $

or $5/4-5\sqrt{2}/4 \le x \le 5/4+5\sqrt{2}/2 $

or $5/4(1-\sqrt{2}) \le x \le 5/4(1+\sqrt{2}) $

and these are implied by the hypotheses.

Answer 2

Continuing from your unpacked conclusion:

Let $f$ be the polynomial

$f(w) = 16w - 40w -25 $, with roots:

$$ w = \frac{5}{4}(1 \pm \sqrt 2)$$.

$f$ obtains negative values between the roots, i.e. in the interval $\left(\>\frac 5 4 (1 - \sqrt 2), \frac 5 4 (1 + \sqrt 2) \> \right)$.

As $$\left( \frac 5 2, \frac 5 4 (1 + \sqrt 2) \right) \subset \left(\>\frac 5 4 (1 - \sqrt 2), \frac 5 4 (1 + \sqrt 2) \> \right)$$

the conclusion follows.