<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] We have
$$1- \cos A = 1-  \frac{b^2+c^2-a^2}{2bc} = \frac{(a+b-c)(c+a-b)}{2bc},$$
so
$$(1-\cos A)(1-\cos B)(1-\cos C) = \frac{(a+b-c)^2(b+c-a)^2(c+a-b)^2}{8a^2b^2c^2},$$
and
$$\cos A \cos B \cos C = \frac{(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)}{8a^2b^2c^2}.$$
Thefore, the original inequality become
$$(a+b-c)^2(b+c-a)^2(c+a-b)^2 \geqslant (a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2).$$
It's remain to prove that
$$(a+b-c)^2(c+a-b)^2 \geqslant (a^2+b^2-c^2)(c^2+a^2-b^2).$$
But, this is true because
$$(a+b-c)^2(c+a-b)^2 - (a^2+b^2-c^2)(c^2+a^2-b^2) = 2(b-c)^2(b^2+c^2-a^2) \geqslant 0.$$
The proof is completed.
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