Question

$0<\alpha <\beta ,\gamma <2\pi$ and $\alpha +\beta +\gamma =\pi$. Prove that $\sum \cos \alpha \ge -\frac{3}{2}$.
Deduce that, in any $△ABC,\sum \cos A\le \frac{3}{2}$.

Answer 1

Let $z=\cos \alpha +\cos \beta +\cos \gamma$ and $\gamma$ be fixed.

Then $z=2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}+\cos \gamma$

$=2\sin \frac{\gamma }{2}\cos \frac{\alpha -\beta }{2}+\cos \gamma$ [as $\frac{\alpha +\beta }{2}=90-\frac{\gamma }{2}$]

$\therefore \gamma$ is constant,$z$ will be maximum when $\cos \frac{\alpha -\beta }{2}$ is maximum,

i.e. $\cos \frac{\alpha -\beta }{2}=1=\alpha =\beta$

Thus,

when angle $gamma$ is fixed, $z$ will be maximum if $\alpha =\beta$

Similarly, angle $beta$ is fixed, $z$ will be maximum if $\gamma =\alpha$

and when angle $gamma$ will be maximum if $\beta =\gamma$

$\therefore z$ will be maximum if $\alpha =\beta =\gamma$

but $\alpha +\beta +\gamma ={180}^{\circ }$

$\therefore \alpha =\beta =\gamma ={60}^{\circ }$

$\therefore z_{max}=\cos {60}^{\circ }+\cos {60}^{\circ }+\cos {60}^{\circ }=\frac{3}{2}$

$\therefore$ In triangle ABC, $\sum \cos A\le \frac{3}{2}$

& $\sum \cos \alpha \ge \frac{-3}{2}$