- #1

anemone

Gold Member

MHB

POTW Director

- 3,883

- 115

By using the Cauchy-Schwarz inequality, we can say that

$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{sinxcosx})^2$ ------(*)

But we know that $ sinxcosx=\frac{sin2x}{2} $

and we're given $ 0 < x < \frac{\pi}{2}$, therefore, $ 0 <sin2x < 1$ and this means $ 0 < \frac{sin2x}{2} < \frac{1}{2}$

From the equation (*), in order to prove $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{sinxcosx})^2$, we need to have a maximum of $ sinxcosx $, and this happens when $ sinxcosx=\frac{1}{2} $.

Now, the inequalities becomes $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{\frac{1}{2}})^2$

$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+2ab)^2$------(**)

Since $ a, b \geq 0 $,

$ a \geq \sqrt a $

$ b \geq \sqrt b $

$ ab \geq \sqrt {ab} $

$ 2ab \geq 2\sqrt {ab} $

$ 1+2ab \geq 1+2\sqrt {ab} $

$ (1+2ab)^2 \geq (1+2\sqrt {ab})^2 $------(***)

From equations (**) and (***), it's obvious that $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+2 \sqrt {ab})^2$.

Am I doing this correct?

Thanks, as usual. :)