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AM,GM,HM Inequality

Prove that: I...

Question

Prove that: If $0 < α, β, γ < \frac{π}{2}$ , prove that $sin α + sin β + sin γ > sin (α + β + γ)$ .

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Solution

Expansion of $sin (α + β + γ)$ we have to prove that $sin α (1 - cos β cos γ) + sin β (1 - cos γ cos α) + sin γ (1 - cos α cos β) + sin α sin β sin γ > 0$
Above is clearly true as $0 < α, β, γ < \frac{π}{2}$ and each of $1 - cos β cos γ$ is also $+$ ive.

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0 < α, β, γ < π / 2

sin α + sin β + sin γ > sin (α + β + γ)

α, β, γ \in (0, \frac{π}{2})

then the value of

\frac{sin (α + β + γ)}{sin α + sin β + sin γ}

α, β, γ ϵ (0, \frac{π}{2})

\frac{sin (α + β + γ)}{sin α + sin β + sin γ}

cos (α - β) + cos (β - γ) + cos (γ - α) = - \frac{3}{2}

then prove that

cos α + cos β + cos γ = sin α + sin β + sin γ = 0

cos α + cos β + cos γ = 0 = sin α + sin β + sin γ

cos 3 α + cos 3 β + cos 3 γ = 3 cos (α + β + γ)

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AM,GM,HM Inequality

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