For $0<x<\dfrac{\pi}{4}$ prove that $$\frac{\cos x}{\sin^2x(\cos x-\sin x)}>8$$

I am trying to use the following for x $\in$ (0,$\frac{\pi}{4}$)

Cos x $\in$ ($\frac{1}{√2}$,1)

(sinx)^2 $\in$ (0,$\frac{1}{2}$)

(cosx-sinx) $\in (0,1)$


Let me try. We have $$\frac{\cos x}{\sin^2 x (\cos x - \sin x)} = \frac{1+\tan^2 x}{\tan^2 x(1-\tan x)}.$$

Note that $0 < x < \dfrac{\pi}{4}$, so $0 < \tan x < 1$,

$$\tan x(1-\tan x) \leq \frac{1}{4}(\tan x + 1-\tan x)^2 = \frac{1}{4}.$$

So we have $$\text{LHS} \geq 4\frac{1+\tan^2x}{\tan x} \geq 8.$$

The equality happens when $\tan x = \dfrac{1}{2}$ for first and $\tan x = 1$ for second, so $\text{LHS} > 8$.

  • 1
    $\begingroup$ can you explain the step $$\tan x(1-\tan x) \leq \frac{1}{4}(\tan x + 1-\tan x)^2 = \frac{1}{4}.$$ $\endgroup$ – Samar Imam Zaidi Aug 28 '17 at 5:44
  • $\begingroup$ It is simple: $(a+b)^2 \geq 4ab$ with $a, b \geq 0$. $\endgroup$ – GAVD Aug 28 '17 at 5:46

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