<s>[INST] Prove |cos(x−1)|+|cos(x)|+|cos(x+1)|≥3/2 I'm working on an induction proof, but I keep coming up against a brick wall.
While working through the induction proof process I keep ending up with $$|\cos(m)|\ge\frac12$$ ,but clearly this isn't true for all m. From another post someone gave me the idea to look at triples and to consider $$|\cos(x−1)|+|\cos(x)|+|\cos(x+1)|$$ to try and find a lower bound. If you plug the above equation into Wolfram Alpha it tells you that $$|\cos(x−1)|+|\cos(x)|+|\cos(x+1)|\ge\frac32$$  Which would prove my induction proof. I just don't know how to go about proving that it does truly exceed $\frac32$
If it helps the original statement that I was trying to prove is $$\sum_{k=0}^n|\cos(k)|\ge\frac n2$$
 [/INST] Well I have a proof for you but its not the most elegant.
Assume first that $\cos(x+1)$ and $\cos(x-1)$ have opposite signs, then 
$$|\cos(x+1)|+|\cos(x-1)|=|\cos(x+1)-\cos(x-1)|=|2\sin (1) \sin x|$$
Now $$|2\sin 1 \sin x+\cos x |\leq |2\sin 1 \sin x|+|\cos x|$$ 
Now a bit of calculus, the minimum of $a\sin x+\cos x$ occurs when $\tan x =a$, so its minimum value is 
$$a\sin x+\cos x=\cos x(a\tan x+1)=\cos x (\tan^2 x+1) =\sec x=\sqrt{a^2+1}$$
Now $\frac{3}{2}\leq \sqrt{a^2+1}$ holds if $\frac{\sqrt{5}}{2} \leq a$ and since $a=2\sin 1$ we have to see that $\frac{\sqrt{5}}{4} \leq \sin 1$. But $\frac{\pi}{4} < 1$ so we are reduced to showing $\frac{\sqrt{5}}{4} \leq \frac{1}{\sqrt{2}}$ which is easy.
Now consider what happens if $\cos(x+1)$ and $\cos(x-1)$ have equal signs. Just to fix ideas lets assume they are positive. Then for an appropriate $n$,
$$-(\frac{\pi}{2}-1) < x+2\pi n < \frac{\pi}{2}-1$$ so in particular
$$\sin 1=\cos (\frac{\pi}{2}-1)\leq \cos x$$.
Now in this case we have 
$$|\cos(x)|+|\cos(x+1)|+|\cos(x-1)|=|\cos(x)+\cos(x+1)+\cos(x-1)|=|\cos(x)(2\cos 1 +1)|$$
Thus with the above inequality we have to show 
$$\frac{3}{2}\leq  |\sin 1(2\cos 1 +1)|.$$
Now 
$\sin 1(2\cos 1 +1)=\sin 2+\sin 1$
using $2 < \frac{2\pi}{3}$ and $\frac{\pi}{4}<1$ we have 
$$\frac{\sqrt{3}}{2}\leq \sin 2$$
$$\frac{\sqrt{2}}{2}\leq \sin 1$$
So we have $$\frac{3}{2}\leq \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}$$ which is easily checked.
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