<s>[INST] Prove $ (\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1$ Let $\triangle ABC$ be a acute triangle. Prove that: $$(\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1 $$
My attempt:
$$\Leftrightarrow (1-cosA)(1-cosB)(1-cosC)\ge cosA.cosB.cosC$$
$$\Leftrightarrow 1-2cosA.cosB.cosC + cosA.cosB + cosB.cosC+cosA.cosC \ge cosA+cosB+cosC $$
$$\Leftrightarrow cos^2A+cos^2B+cos^2C + cosA.cosB + cosB.cosC+cosA.cosC \ge cosA+cosB+cosC $$
$$ 0<cos A,cos B,cosC<1$$
$$cos^2A+cos^2B+cos^2C=1-2cosA.cosB.cosC\ge 3\sqrt[3]{cosA.cosB.cosC}$$
$$\Rightarrow cosA.cosB.cosC \le \frac{1}{8}$$
And I was stuck here. Could you help me ?
 [/INST] Simple. Use AM-GM and Jensen's inequalities: $\cos A\cos B\cos C \le \dfrac{(\cos A+\cos B+\cos C)^3}{27}\le \dfrac{1}{27}\cdot (3\cos (\frac{A+B+C}{3}))^3=\left(\frac{1}{2}\right)^3 =\dfrac{1}{8}.$
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