I am trying to prove that $$\cos(a)+\cos(b)-\cos(a+b)\geq 1$$ For $a,b \geq 0$ and $0\leq a+b\leq 180^°$
I have checked in Wolfram Alpha that the inequality is true, but I am not able to prove it. The trigonometric identitiy I have tried to apply (basically, $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b))$$ does not seem useful as it is, so if there is some other one that you think it could help… any hint would be welcomed.
Thanks!
Best Answer
$$\cos a + \cos b - \cos(a+b) - 1 = 2\cos\frac{a+b}{2}\cos\frac{a-b}{2} - 2\cos^2\frac{a+b}{2} = 2\cos\frac{a+b}{2}\left(\cos\frac{a-b}{2} - \cos\frac{a+b}{2}\right)\geq 0.$$ The thing in the parenthesis is non-negative because $\cos$ decreases on $[0,\pi/2].$