Prove that for all real $\displaystyle x$ and $\displaystyle y$
$\displaystyle
|\cos x|+|\cos y|+|\cos(x+y)| \geq 1
$
Notice first that $\displaystyle |\cos x| = |\cos(\pi+x)| = |\cos(\pi-x)|$. 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 $\displaystyle x+y>\pi$ then we can replace x by $\displaystyle \pi-x$ and y by $\displaystyle \pi-y$. That will have the effect of replacing x+y by $\displaystyle 2\pi-(x+y)$, and again the values of $\displaystyle |\cos x|$, $\displaystyle |\cos y|$ and$\displaystyle |\cos (x+y)|$ will be unchanged. Therefore we can assume that $\displaystyle x+y<\pi$ and hence x, y and $\displaystyle \pi-(x+y)$ are the three angles of a triangle.
So we want to prove that $\displaystyle |\cos A| + |\cos B| + |\cos C|\geqslant1$, where A, B, C are the angles of a triangle with sides a, b, c. In fact, the stronger inequality $\displaystyle \cos A + \cos B + \cos C > 1$ is true.
The cosine rule says that $\displaystyle \cos A + \cos B + \cos C = \dfrac{b^2+c^2-a^2}{2bc} + \dfrac{c^2+a^2-b^2}{2ca} + \dfrac{a^2+b^2-c^2}{2ab} = \dfrac{a^2(b+c) + b^2(c+a) + c^2(a+b) - a^3-b^3-c^3}{2abc}.$
So we want to prove that $\displaystyle a^2(b+c) + b^2(c+a) + c^2(a+b) - a^3-b^3-c^3 > 2abc$. But we know that $\displaystyle b+c>a$, $\displaystyle c+a>b$ and $\displaystyle a+b>c$ and therefore $\displaystyle (b+c-a)(c+a-b)(a+b-c)>0$. If you multiply that out, you will find that it gives exactly the inequality that is wanted.