11RHS is trivially proved by Jenson's Inequality [11]
in any tri.... ABC prove that...
1 < cos A + cos B + cos C <= 3/2
<= means less than or equal to
nd ya.... thanks in advance ....
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9 Answers
21cosA + cosB + cosC - 32≤0
<=> 2sin2C2 - 4sinC2cos(A-B)2 + 1 ≤0
the expression is a quadratic in sin(C2) and we want its discriminant ≤0
which is equivalent to cos2(A-B)2≤1 which is true hence proved
try the LHS
3cosA + 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.
21yes can be proved but not toooo trivially
any way please show ur method..[1]
11\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.
11To 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]
21I think they gave 1 because
1<1+ \frac{r}{R}= cosA + cosB + cosC
Using R\geq 2r we can probe the RHS as well
3devil is correct.
cause 1 + 4sinA2sinB2sinC2
directly implies this....
this has to greater than 1 cause A + B + C = pi.
and it is a triangle.
