Olympiad Maths Prep

Track / Stage 7 / 72 of 300 #1472 of 2000

Problem 1472

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.2 Prove it

□\square Example 5α,β,γ5 \alpha, \beta, \gamma are the three interior angles of a given triangle. Prove that csc⁡2α2+\csc ^{2} \frac{\alpha}{2}+ csc⁡2β2+csc⁡2γ2⩾12\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 12, and find the condition for equality. (1994 National Mathematical Olympiad Problem)

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Prove that by the AM-GM inequality,
csc⁡2α2+csc⁡2β2+csc⁡2γ2⩾3csc⁡2α2csc⁡2β2csc⁡2γ23\csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 3 \sqrt[3]{\csc ^{2} \frac{\alpha}{2} \csc ^{2} \frac{\beta}{2} \csc ^{2} \frac{\gamma}{2}}

where equality holds if and only if α=β=γ\alpha=\beta=\gamma. Then, by the AM-GM inequality and the concavity of the sine function, we have
sin⁡α2sin⁡β2sin⁡γ23⩽sin⁡α2+sin⁡β2+sin⁡γ23⩽sin⁡α2+β2+γ23=sin⁡α+β+γ6=12,csc⁡2α2+csc⁡2β2+csc⁡2γ2⩾12,\begin{array}{l} \sqrt[3]{\sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2}} \leqslant \frac{\sin \frac{\alpha}{2}+\sin \frac{\beta}{2}+\sin \frac{\gamma}{2}}{3} \\ \leqslant \sin \frac{\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}}{3}=\sin \frac{\alpha+\beta+\gamma}{6}=\frac{1}{2}, \\ \csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 12, \end{array}

Therefore,
where equality holds if and only if α=β=γ\alpha=\beta=\gamma.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.