Prove Trig. Inequality: A,B & C Are Triangle Angles

[Jean-Louis]

Show that :

(sinA/2 + sinB/2 + sinC/2)^2 >= (sinA)^2 + (sinB)^2 + (sinC)^2

A,B and C are the angles of a triangle.

 

[Gib Z]

Are you looking for an elementary proof using trig manipulations or identites?
I can't see anything, Ill just right the question again that other people might find easier to help you with.

Show that \frac{\sin A + \sin B + \sin C}{2} >= \sin^2 A + \sin^2 B + \sin^2 C
bounded by the condition A+B+C=\pi

 

[murshid_islam]

Gib Z, i think you have wrote the question wrong. i think Jean-Louis wrote,

\left(\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2}\right)^{2} \geq \sin^2 A + \sin^2 B + \sin^2 C where A + B + C = \pi

 

[cristo]

Show that :

(sinA/2 + sinB/2 + sinC/2)^2 >= (sinA)^2 + (sinB)^2 + (sinC)^2

A,B and C are the angles of a triangle.

Have you tried doing anything? I'd multiply out the left hand side first, and see what happens.

 

[Jean-Louis]

I think I got it. Here it goes :

sin(x/2) = +/- sqrt((1-cos(x))/2)
and sin^2(z) = 1 - cos^2(x)

so that that (sinA)^2 + (sinB)^2 + (sinC)^2 = 3 - (cos^2(A) + cos^2(B) +
cos^2(C))

Now (sin A/2 + sin B/2 + sin C/2)^2 = (+/- sqrt((1-cos(A))/2) + +/-
sqrt((1-cos(B))/2) + +/- sqrt((1-cos(C)/2) ) ^2

If we let

A = +/- sqrt((1-cos(A))/2)
B = +/- sqrt((1-cos(B))/2)
C = +/- sqrt((1-cos(C))/2)

Then (sin A/2 + sin B/2 + sin C/2)^2 = A^2 + B^2 + C^2 + 2AB + 2AC + 2BC

A^2 = (1 - cos(A))/2
B^2 = (1 - cos(B))/2
C^2 = (1 - cos(C))/2

AB = 1/2 SQRT((1-cosA)(1-cosB))
AC = 1/2 SQRT((1-cosA)(1-cosC))
BC = 1/2 SQRT((1-cosB)(1-cosC))

so you have

(1 - cos(A))/2 + (1 - cos(B))/2 + (1 - cos(C))/2 +
SQRT((1-cosA)(1-cosB)) + SQRT((1-cosA)(1-cosC)) + SQRT((1-cosB)(1-cosC))
>= 3 - (cos^2(A) + cos^2(B) + cos^2(C))

Now the Law of cosines tells us this...

cosC = (a^2 + b^2 -c^2)/2ab
cosA = (b^2 + c^2 - a^2)/2bc
cosB = ((a^2 + c^2 - b^2)/2ac

You have all the equations in terms of cosines - substitute simplify...

QED

 

[Gib Z]

Kill me. Somebody kill me...