Prove the inequality $(a+b)(b+c)(a+c)\ge8$ if $a^2+b^2+c^2+abc=4$.

Question

I noted $a=2\cos A,b=2\cos B,c=2\cos C$ because of the identity $\cos^2(x)+\cos^2(y)+\cos^2(z)+2\cos(x)\cos(y)\cos(z)=1$,but the calculations looked really bad, so I gave up. Any another suggestion?

Answer 1

You're on the right track in terms of strategy, but the stated inequality is the wrong way round: for example, an isosceles right-angled triangle gives$$a=b=\sqrt{2},\,c=0\implies(a+b)(b+c)(c+a)=4\sqrt{2}<8.$$With your parameterization,$$a+b=2(\cos A+\cos B)=4\cos\tfrac{A+B}{2}\cos\tfrac{A-B}{2}=4\sin\tfrac{C}{2}\cos\tfrac{A-B}{2}$$etc., so$$(a+b)(b+c)(c+a)=64\sin\tfrac{A}{2}\sin\tfrac{B}{2}\sin\tfrac{C}{2}\underbrace{\cos\tfrac{A-B}{2}\cos\tfrac{B-C}{2}\cos\tfrac{C-A}{2}}_{\in[0,\,1]}.$$Applying Jensen's inequality to $\ln\sin\tfrac{x}{2}$ implies $\sin\tfrac{A}{2}\sin\tfrac{B}{2}\sin\tfrac{C}{2}\le\tfrac18$, achieving its maximum for an equilateral triangle.