For non-negative $a$ and $b$ with $a+b \leq c$ for a small constant $c$, what is the minimum of $\cos a + \cos b$? 
Let $a,b \geq 0$ with $a+b \leq c$ for a small constant $c$ between $0$ and $1$.
What is the minimum of $\cos(a) + \cos(b)$?

I conjecture it is $\cos(0)+\cos(c) = 1 + \cos(c)$ but I have no proof for this.

$\cos x$ is strictly decreasing and positive for $x\in [0,1].$ So $\cos a +\cos b$ cannot be a minimum if $a<c-b$ because then $\cos a +\cos b >\cos (c-b)+\cos b.$
For $a=c-b$ let $M=(a+b)/2=c/2$ and $D=(a-b)/2.$ We have $$\cos a=\cos (M+D)=\cos M \cos D-\sin M\sin D$$ and  we have $$\cos b=\cos (M-D)=\cos M \cos D+\sin M \sin D.$$ Adding, we have (since $\cos M$ and $\cos |D|$ are positive) $$\cos a +\cos b=2\cos M\cos D=$$ $$=2\cos M \cos |D| \ge$$ $$\ge 2\cos M\cos (c/2)=$$ $$=2\cos^2 (c/2)=1+\cos c$$ with equality throughout iff $|D|=c/2$, that is, iff $\{a,b\}=\{0,c\}.$
Footnote: For a proof that $\cos x$ is strictly decreasing for $x\in [0,1]$: If $0\le x<y\le 1$ let $m=(y+x)/2$ and $d=(y-x)/2.$ Then $$\cos y-\cos x =\cos (m+d)-\cos (m-d)=$$ $$=(\cos m \cos d-\sin m \sin d)-(\cos m \cos d+\sin m \sin d)=$$ $$=-2\sin m\sin d<0$$ because $0<d<m<1$ so $\sin m$ and $\sin d$ are positive.
