By considering "Arithmetic mean ≧ Geometric mean" Prove the following inequality: sinA + sinB + sinC ≦ 3(sqrt3) /2 where A+B+C=180°Please do not prove the inequality by using others like...

By considering "Arithmetic mean ≧ Geometric mean"

Prove the following inequality:

sinA + sinB + sinC ≦ 3(sqrt3) /2

where A+B+C=180°

Please do not prove the inequality by using others like Jensen's inequality as the question require us to prove that inequality by  using "A.M. ≧ G.M."

Asked on by alan0552

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neela | High School Teacher | (Level 3) Valedictorian

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Angles A+B+C =180

To prove: sinA+sinB+sinC = 3 *(sqrt3)/2

Solution:

sinA+sinB+sinC = sinA+(sinB+sinC)=sinA+2sin((B+C)/2) * cos ((B-C)/2)

= 2sin(A/2) *cos(A/2)+2 sin(90-A/2)*cos((B-C)/2)

=2sin(A/2) *cos(A/2)+2 cos(A/2)*cos((B-C)/2)

=2cos(A/2){sin(A/2)+cos((B-C)/2)}

=2cos(A/2){cos(B+C)/2)+cos((B-C)/2)}

=4cos(A/2)cos(B/2)(cos(C/2)

= 4 {GM of cos(A/2),cos(B/2), and (cos(C/2)]}^3

<=4 {(1/3)AM of[ cos(A/2),cos(B/2), and (cos(C/2)}^3.

=4{(1/3)[ cos(A/2)+cos(B/2)+(cos(C/2)}^3......................(1)

But GM  of x1, x2, x3,....xn, could be equal to AM , if and only if  there is no variations between x's . In other words,all x's are equal , that is, x1=x2=x3=.....=xn, when the value of GM attains maximum and is equal to AM.

So, for maximum value of GM, here, cos(A/2) =cos(B/2) = cos(C/2) , which is possible if and only if A/2 = B/2 =C/2 = [(A+B+C)/2]/3 =30.

So on the right side value  of the inequality at (1) , is as below:

sinA+sinB+sinC <= 4{(1/3)[ (sqrt3)/2 +(sqrt3)/2 +(sqrt3)/2 }^3

=4*{(1/3)*3*(sqrt3/2)}^3

=4{(sqrt3)/2}^3

=3*(sqrt3)/2.

****************

Aliter:

Since, A+B+C =180 degrees, A, B and C are the vertices of a triangle.The sine relation of the sides and angles of the triangle is as follows.

Area Delta (or D) of the triangle with usual notations is given by:

D = (1/2) bc sinA, giving,

sinA = 2aD /abc

sinB =2bD/abc and

sinC = 2c/abc. Therefore.

Sina+sinB+sinC =2D(a+b+c) /abc.

=2*2s[s(s-a)(s-b)(s-c)]^(1/2)/abc , by Heron's formula, where  2s=a+b+c

=2*2s*s^(1/2){[(s-a)/a^2]*(s-b)/b^2][s-c)/c^2}^(1/2)

=2*2s*s^(1/2){{Geometric mean of [(s-a)/a^2]), {(s-b)/b^2] and[[s-c)/c^2])} ^3 )^(1/2)

<= 22s*s^(1/2) {(1/3)Arithmetic mean of {[(s-a)/a^2]), {(s-b)/b^2] and[[s-c)/c^2])} ^3 }^(1/2)

=4s*s^(1/2){(1/3)[(s-a)/a^2])+ {(s-b)/b^2]+[[s-c)/c^2])} ^3 )^(1/2)...................(1)

The GM of x1,x2 and x3,...xn  is less than their AM when x1 and x2 ,x3, ...xn and GM = AM when x1 = x2=x3= ...=xn or all of them are equal. So the GM attains maximun at this event.

Therefore, (s-a)/a^2 =(s-b)/b^2=(s-c)/c^2 .This needs  a =b=c = a and s= 3a/2 The expresson at (1) becomes:

6a*(3a/2)^(1/2).{(1/3)[(1/(2a)+1/(2a)+1/(2a)]^3/2

=6a^(3/2)* (3/2)^(1/2)*{(1/(2a)^(1/3)}

=3*[3^(1/2)]/2

Therefore,

sinA+sinB+sinC <= 3(sqrt3)/2

 

 

 

 

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