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?
Asked
Active
Viewed 63 times
-1
-
1Typo: do you mean $...\gt 8$ if $a^2+..$? – herb steinberg Mar 16 '21 at 21:58
-
1Not directly on point, but closely related: see this aops article. – user2661923 Mar 17 '21 at 17:53
1 Answers
2
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.
J.G.
- 115,835