IIT JEE , AIEEE Forum - Mathematics , Trignometry

ramkumar_november

try this problem ...

$If\;\; 0<x<\frac{\pi}{4}\; ,\;\; then\;\; prove\;\; that$

$\frac{cosx}{sin^2x(cosx-sinx)}>8$

hsbhatt

$\frac{\cos x}{\sin^2 x (\cos x - \sin x)}$

= $\frac{1}{\sin^2 x (1-\tan x)}$

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

Hence the inequality becomes

$\frac{1+\tan^2 x}{\tan^2 x (1-\tan x)} \ge 8$

or $8\tan^3 x + 1 \ge 7 \tan^2x$ (since 1-tanx > 0)

$LHS = 8\tan^3 x+1 = 4\tan^3x + 4\tan^3 x + 1 \ge 3 \sqrt [3] {16\tan^6 x}$

= $6 \sqrt [3] {2} \tan^2 x$

It only remains to prove that $6 \sqrt [3] {2} \tan^2 x \ge 7 \tan^2 x$

or $6 \sqrt [3] {2} \ge 7$ which is found to be true by cubing both sides.

So, the inequality is a strict inequality