cos a + cos b + cos c hmm....

cosA + cosB + cosC

or, 2cos(A+B)2 cos(A-B)2 + cosC

also A + B + C = 180 degrees.
so, cos(A +B)2 = cos(90 - C2) = sinC2
so we get,
2sinC2cos(A-B)2 + cosC
or, 2sinC2cos(A-B)2 + 1 - 2sin²C2
or, 1 + 2sinC2 { cos(A-B)2 - sinC2}
but the maximum value of cos(A-B)2 is 1.
so the above obtained equation is less than or equal to
1 + 2sinC2{1 - sinC2}
or, 1 + 2sinC2 - 2sin²C2
or, 1 - 2[sin²C2 - sinC2 + 14 - 14]
or, 1 - 2[sinC2 - 12]² + 24
32 -2[sinC2 - 12]²

so the maximum value is 32.

yes can be proved but not toooo trivially

any way please show ur method..[1]

\frac{\cos A+\cos B}{2}\le \cos \frac{A+B}{2}=\sin \frac{C}{2}

That gives \sum{\cos A}\le \sum{\sin \frac{A}{2}}\le \frac{3}{2} by Jenson's.

To elaborate a bit, this is the std. way we prove the following inequality:

1<\sum{\left(\frac{\sin A-\sin B}{A-B} \right)}\le \frac{3}{2}

In-fact this itself uses the inequality stated. Also 1 is not a proper lower bound for the inequality I stated. A tighter bound is \frac{4}{\pi}

[1]

devil is correct.
cause 1 + 4sinA2sinB2sinC2
directly implies this....
this has to greater than 1 cause A + B + C = pi.
and it is a triangle.