# Nice Trigonometric inequality

Prove that for all real numbers x,y, |cos(x)|+ |cos(y)| + |cos(x+y)|\geq 1

• Shubhodip ·

Hint: 1= cos(x+y-(x+y)) = .....â‰¤....â‰¤....

• Hari Shankar ·

WLOG |\cos x| \ge |\cos y| so that it follows that |\sin x| \le |\sin y|

Now,

|\cos x| + |\cos y| + |\cos(x+y)| = |\cos x| + |\cos y| + |\cos x \cos y - \sin x \sin y|

\ge |\cos x| + |\cos y| - |\cos x| \cos y| + |\sin x| |\sin y|

= |\cos y|(1-|\cos x|) + |\cos x| + |\sin x| |\sin y|

\ge |\cos y|(1-|\cos x|) + \cos^2 x + \sin^2 x \ge 1

• Shubhodip ·

oh sorry its perfectly correct ...i was flipping inequalities in mind

|a|+|b|â‰¥|a-b|â‰¥|a|-|b| is correct i'm fine

deleted my post to keep things clean

• Shubhodip ·

Somehow ordering did not come in my mind.

I did this . Do tell me if it is wrong.

1= |cos^2(x+y) + sin^2(x+y)| \leq |cos(x+y)| + |sin(x+y)| \leq |cos(x+y)|+ |sin x cos y+ sin y cos x |\leq |cos(x+y)|+ |sin x||cos y| + |cos x||sin y| \leq |cos(x+y)| + |cos x| + |cos y|

• Hari Shankar ·

That is correct shubhodip.

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