Prove $ (\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1$

Question

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 ?

Answer 1

Simple. Use AM-GM and Jensen's inequalities: $\cos A\cos B\cos C \le \dfrac{(\cos A+\cos B+\cos C)^3}{27}\le \dfrac{1}{27}\cdot (3\cos (\frac{A+B+C}{3}))^3=\left(\frac{1}{2}\right)^3 =\dfrac{1}{8}.$

Answer 2

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.