1. ## Trigonometric Inequality

Prove that for all real $x$ and $y$

$
|\cos x|+|\cos y|+|\cos(x+y)| \geq 1
$

2. A few things you should know:

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

$|\cos x| \leq 1$

$|\sin x| \leq 1$

3. Originally Posted by abender
A few things you should know:

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

$|\cos x| \leq 1$

$|\sin x| \leq 1$
Hello abender,

I am curious to know how these simple identities will help. Can you give us some more hints?

4. Originally Posted by pankaj
Prove that for all real $x$ and $y$

$
|\cos x|+|\cos y|+|\cos(x+y)| \geq 1
$
Notice first that $|\cos x| = |\cos(\pi+x)| = |\cos(\pi-x)|$. So by adding multiples of pi to x and y, we may assume that both x and y lie between 0 and pi. Also, if $x+y>\pi$ then we can replace x by $\pi-x$ and y by $\pi-y$. That will have the effect of replacing x+y by $2\pi-(x+y)$, and again the values of $|\cos x|$, $|\cos y|$ and $|\cos (x+y)|$ will be unchanged. Therefore we can assume that $x+y<\pi$ and hence x, y and $\pi-(x+y)$ are the three angles of a triangle.

So we want to prove that $|\cos A| + |\cos B| + |\cos C|\geqslant1$, where A, B, C are the angles of a triangle with sides a, b, c. In fact, the stronger inequality $\cos A + \cos B + \cos C > 1$ is true.

The cosine rule says that $\cos A + \cos B + \cos C = \dfrac{b^2+c^2-a^2}{2bc} + \dfrac{c^2+a^2-b^2}{2ca} + \dfrac{a^2+b^2-c^2}{2ab} = \dfrac{a^2(b+c) + b^2(c+a) + c^2(a+b) - a^3-b^3-c^3}{2abc}.$

So we want to prove that $a^2(b+c) + b^2(c+a) + c^2(a+b) - a^3-b^3-c^3 > 2abc$. But we know that $b+c>a$, $c+a>b$ and $a+b>c$ and therefore $(b+c-a)(c+a-b)(a+b-c)>0$. If you multiply that out, you will find that it gives exactly the inequality that is wanted.