Let-a-b-gt-0-and-x-0-pi-2-Prove-a-sinx-1-b-cosx-1-1-2ab-2-

Question Number 117832 by snipers237 last updated on 13/Oct/20

L e t a, b > 0 a n d x \in] 0; \frac{π}{2} [

P r o v e (\frac{a}{s i n x} + 1) (\frac{b}{c o s x} + 1) ⩾ {(1 + \sqrt{2 a b})}^{2}

Answered by 1549442205PVT last updated on 14/Oct/20

From the hypothesis a, b > 0 a n d x \in] 0; \frac{π}{2} [

we have sinx > 0, cosx > 0 . Hence

L . H . S = \frac{a}{sinx} + \frac{b}{cosx} + \frac{ab}{sinxcosx} + 1

= {(\sqrt{\frac{a}{sinx}} - \sqrt{\frac{b}{cosx}})}^{2} + 2 \sqrt{\frac{ab}{sinxcosx}}

+ \frac{2 ab}{\sin 2 x} + 1 ⩾ 2 \sqrt{\frac{2 ab}{\sin 2 x}} + \frac{2 ab}{\sin 2 x} + 1

⩾ 2 ab + 2 \sqrt{2 ab} + 1 = {(1 + \sqrt{2 ab})}^{2}

(due to 0 < \sin 2 x ⩽ 1) . Hence, the inequality

(\frac{a}{s i n x} + 1) (\frac{b}{c o s x} + 1) ⩾ {(1 + \sqrt{2 a b})}^{2}

is proved . The equality ocurrs if and

only if

{\begin{cases} \frac{a}{sinx} = \frac{b}{cosx} \\ \sin 2 x = 1 \end{cases} \Leftrightarrow {\begin{cases} x = \frac{π}{4} \\ a = b \end{cases}