Question

For a triangle ABC, prove that
$cosA+cosB+cosC\le 3/2.$
In case of equility, triangle will be equilateral.

Answer 1

$E=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2si{n}^2\frac{C}{2}-\frac{3}{2}$
=$-2si{n}^2\frac{C}{2}+2cos\frac{A-B}{2}sin\frac{C}{2}-\frac{1}{2}$
=$-2x^2+2xcos\frac{A-B}{2}-\frac{1}{2},x=sin\frac{C}{2}$ ..............(1)
Now we know that sign of a quadratic is the same as that of the first term provided $△<0$.
Here $△=4co{s}^2\frac{A-B}{2}-4$, which is clearly -ive as
$co{s}^2\frac{A-B}{2}\le 1$
$\therefore E\le 0$.
$\therefore cosA+cosB+cosC\le 3/2$
Equality will be possible when $co{s}^2\frac{A-B}{2}=1$
(i.e. $△$=0) or A=B.
Then from (1),
$4x^2-4x+1=0$ or $(2x-1{)}^2=0$
$\therefore x=sin\frac{C}{2}=\frac{1}{2}\therefore C={60}^0\therefore A=B=C={60}^0$
Hence the triangle is equilateral.