the problem may be wrong,let $x=y=\dfrac{\pi}{3}, z=-\dfrac{2\pi}{3}$,there is a contradiction!

suweijie wrote: the problem may be wrong,let $x=y=\dfrac{\pi}{3},z=-\dfrac{2\pi}{3}$,there is a contradiction! You are right Thanks, I have corrected the statement now

let me have a try.put $a=|\cos{x}|,b=|\cos{y}|$,then $|\cos{z}|=|\cos(x+y)|=|\cos{x}\cos{y}-\sin{x}\sin{y}|\geq |ab-\sqrt{(1-a^2)(1-b^2)}|$.define $s=\arccos{a},t=\arccos{b}$,thus $|ab-\sqrt{(1-a^2)(1-b^2)}|=|\cos(s+t)|$.Since $s,t$ are in the interval $[0,\dfrac{\pi}{2}]$,so if $s+t \le \dfrac{\pi}{2}$,then $\cos{s}+\cos{t}+|\cos(s+t)|\geq\cos{s}+\cos{t}\geq\cos^2s+\cos^2t\geq 1$. if $s+t\geq\dfrac{\pi}{2}$, then \begin{eqnarray*} \cos{s}+\cos{t}+|\cos(s+t)| &=& \cos{s}+\cos{t}-\cos(s+t) \\ \ &=& 4\sin\dfrac{s}{2}\sin\dfrac{t}{2}\cos{\dfrac{s+t}{2}}+1\\ \ &\geq& 1 . \end{eqnarray*}

We have two useful ineqs here : $|sin(x+y)| \leq min(|cosx|+|cosy| ,|sinx|+|siny|)$ $|cos(x+y)| \leq min(|sinx|+|cosy| ,|siny|+|cosx|)$ Using these ineqs , we can prove : $|cosx_1|+|cosx_2|+...+|cosx_{2k+1}| \geq 1$ , where $x_1+x_2+...+x_{2k+1} = 0$

i think it can be proved without any difficulty take for example x=min(x,y,z) we have |cos x |+ |cos y| +| cos z | >= |cos x | x min so 0>=3x so 0>=x so cosx>=cos(0)=1 so it's proven

nawram88 wrote: 0>=x so cosx>=cos(0)=1 obviously false.