Prove that \(\displaystyle{\sin{{\frac{{{A}}}{{{2}}}}}}{\sin{{\frac{{{B}}}{{{2}}}}}}{\sin{{\frac{{{C}}}{{{2}}}}}}\le{\frac{{{1}}}{{{8}}}}\)

aanvarendbq28

aanvarendbq28

Answered question

2022-03-29

Prove that sinA2sinB2sinC218

Answer & Explanation

Madilyn Shah

Madilyn Shah

Beginner2022-03-30Added 11 answers

Note that sinx is concave function when 0xπ2. Then by Jensen's inequality on concave function, we have:
sinA2+sinB2+sinC23sinA+B+C23=sinπ6=12
Now by AM-GM inequality
sinA2+sinB2+sinC23(sinA2sinB2sinC2)13
Hence
sinA2sinB2sinC218
Equality holds when A=B=C=π3
zevillageobau

zevillageobau

Beginner2022-03-31Added 13 answers

This question is the same as asking: when α+β+γ=π2, what is the maximum of sin(α)sin(β)sin(γ)?
We wish to find α,β,γ so that for each δα,δβ,δγ so that
δα+δβ+δγ=0
we also have
0=δsin(α)sin(β)sin(γ)
=cos(α)sin(β)sin(γ),δα+sin(α)cos(β)sin(γ),δβ+sin(α)sin(β)cos(γ),δγ
Since each δα,δβ,δγ that is perpendicular to (1,1,1), it must also be perpendicular to (cot(α),cot(β),cot(γ)), we must have that cot(α)=cot(β)=cot(γ). Thus, α=β=γ=π6 and
sin(α)sin(β)sin(γ)=18

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