MHB Geom. Challenge: Prove $(1-\cos A)(1-\cos B)(1-\cos C)\ge \cos A\cos B \cos C$

[anemone]

Let $A,\,B$ and $C$ be three angles of a triangle $ABC$. Prove that

$(1-\cos A)(1-\cos B)(1-\cos C)\ge \cos A\cos B \cos C$

 

[MarkFL]

Let $A,\,B$ and $C$ be three angles of a triangle $ABC$. Prove that

$(1-\cos A)(1-\cos B)(1-\cos C)\ge \cos A\cos B \cos C$

My solution:

Spoiler

Consider the objective function:

$$f(A,B,C)=\left(1-\cos(A)\right)\left(1-\cos(B)\right)\left(1-\cos(C)\right)-\cos(A)\cos(B)\cos(C)$$

Subject to the constraint:

$$g(A,B,C)=A+B+C-\pi=0$$ where $$0<A,B,C<\pi$$

Now, by cyclic symmetry, we see that the extremum occurs for:

$$A=B=C=\frac{\pi}{3}$$

And we then find:

$$f\left(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}\right)=\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^3=0$$

To show this extremum is a minimum, let's choose another point on the constraint:

$$(A,B,C)=\left(\frac{\pi}{2},\frac{\pi}{4},\frac{\pi}{4}\right)$$

And we find:

$$f\left(\frac{\pi}{2},\frac{\pi}{4},\frac{\pi}{4}\right)=\frac{1}{2}-0=\frac{1}{2}>0$$

Hence, we may conclude:

$$f_{\min}=f\left(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}\right)=0$$

And thus, the inequality is confirmed.

 

[anemone]

My solution:

Spoiler

Consider the objective function:

$$f(A,B,C)=\left(1-\cos(A)\right)\left(1-\cos(B)\right)\left(1-\cos(C)\right)-\cos(A)\cos(B)\cos(C)$$

Subject to the constraint:

$$g(A,B,C)=A+B+C-\pi=0$$ where $$0<A,B,C<\pi$$

Now, by cyclic symmetry, we see that the extremum occurs for:

$$A=B=C=\frac{\pi}{3}$$

And we then find:

$$f\left(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}\right)=\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^3=0$$

To show this extremum is a minimum, let's choose another point on the constraint:

$$(A,B,C)=\left(\frac{\pi}{2},\frac{\pi}{4},\frac{\pi}{4}\right)$$

And we find:

$$f\left(\frac{\pi}{2},\frac{\pi}{4},\frac{\pi}{4}\right)=\frac{1}{2}-0=\frac{1}{2}>0$$

Hence, we may conclude:

$$f_{\min}=f\left(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}\right)=0$$

And thus, the inequality is confirmed.

Very well done, MarkFL!:)

I still welcome those who have a thing for geometry approach to tackle this challenge!

 

[anemone]

Solution of other:

Spoiler

If the triangle $ABC$ is an obtuse triangle, then we're done.

If the triangle $ABC$ is acute, then we see that we have:

$$\prod_{}^{}\cos A\le \prod_{}^{} (1-\cos A)$$ which gives

$$\prod_{}^{}\cos A(1+\cos A)\le \prod_{}^{} (1-\cos^2 A)$$

$$\prod_{}^{}\cos A(1+2\cos^2 \frac{A}{2}-1)\le \prod_{}^{} \sin^2 A$$

$$8\prod_{}^{}\cos A(\cos^2 \frac{A}{2})\le \prod_{}^{} \sin^2 A$$

$$8\cos A\cos B\cos C\cos^2 \frac{A}{2}\cos^2 \frac{B}{2}\cos^2 \frac{C}{2}\le \sin^2 A \sin^2 B \sin^2 C$$

$$8\cos A\cos B\cos C\cos^2 \frac{A}{2}\cos^2 \frac{B}{2}\cos^2 \frac{C}{2}\le \sin A \sin B \sin C \left(2\sin \frac{A}{2} \cos \frac{A}{2}\right)\left(2\sin \frac{B}{2} \cos \frac{B}{2}\right)\left(2\sin \frac{C}{2} \cos \frac{C}{2}\right)$$

$$\prod_{}^{}\cot \frac{A}{2}\le \prod_{}^{} \tan A$$

$$\implies \sum_{}^{} \cot \frac{A}{2}\le \sum_{}^{} \tan A$$ which is certainly true.

Therefore we can conclude by now that our assumption is correct and we've proved that

$(1-\cos A)(1-\cos B)(1-\cos C)\ge \cos A\cos B \cos C$.