Difficult trigonometric inequality

Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot

Difficult trigonometric inequality

Postby Math Tutor » Thu Jan 03, 2008 9:47 am

Prove that:
[tex](1 + \frac{1}{sin(\alpha )})(1 + \frac{1}{cos(\alpha )})>5[/tex]

[tex]\alpha \in (0, 90^\circ)[/tex]
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Re: Difficult trigonometric inequality

Postby dduclam » Sun May 04, 2008 4:53 am

teacher wrote:Prove that:
[tex](1 + \frac{1}{sin(\alpha )})(1 + \frac{1}{cos(\alpha )})>5[/tex]

[tex]\alpha \in (0, 90^\circ)[/tex]


I don't think that it's so hard. Because we can solve a stronger problem:

[tex](1 + \frac{1}{sin(\alpha )})(1 + \frac{1}{cos(\alpha )})>3+2\sqrt2[/tex]

<=> [tex]1+\frac1{sin\alpha}+\frac1{cos\alpha}+\frac1{sin\alpha.cos\alpha}>3+2\sqrt2[/tex] (1)

By AM-GM inequality we have [tex]\frac1{a}+\frac1{b}\ge\frac4{a+b}[/tex]

Then [tex]LSH (1)\ge 1+\frac4{sin\alpha+cos\alpha}+\frac1{sin\alpha.cos\alpha}[/tex]

Obvious: [tex]sin\alpha+cos\alpha\le \sqrt2[/tex] and [tex]sin\alpha.cos\alpha\le\frac1{2}[/tex]

So [tex]LSH (1)\ge 1+\frac4{\sqrt2}+2=3+2\sqrt2>5[/tex] (q.e.d)

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