Prove that $\cos(2a) + \cos(2b) + \cos(2c) \geq -\frac{3}{2}$ for angles of a triangle

Question

Let the three internal angles of a triangle are $a,b,c$. Prove that

$$\cos(2a) + \cos(2b) + \cos(2c) \geq -\frac{3}{2}.$$

I'm looking for an elementary, geometric proof. So avoid derivatives and tools from optimalization theory if it's possible.

Answer 1

Writing $\cos 2x = 1 -\sin^2 x$ and using the sine rule $\frac{a}{\sin \alpha} = 2R$, the inequality turns out to be equivalent to $$ a^2 + b^2 + c^2 \leq 9R^2 $$ where $a,b,c$ are the sides of the triangle and $R$ is the circumradius.

This inequality is sometimes known as Leibniz's inequality.

For this inequality, a geometric proof is possible. Let $O$ and $G$ be the circumcenter and centroid of the triangle, respectively. Let $A$ be one of the vertices and let $M$ be the midpoint of the side $BC$ opposite to $A$. Applying Stewart's theorem in triangles $ABC$ and $AOM$, one can show that $$ |OG|^2 = R^2 - \frac19(a^2+b^2+c^2), \qquad (\ast) $$ thus proving the inequality.

An equivalent form of $(\ast)$ is $|OH|^2 = 9R^2 - a^2-b^2-c^2$, where $H$ is the orthocenter. This equality can be proven using complex numbers and surely also by purely geometric means, for instance by computing the power of $H$ with respect to the circumcircle.

Answer 2

Combine $$\cos(2a)+\cos(2b)+\cos(2c)=-4\cos(a)\cos(b)\cos(c)-1$$ and $$\cos(a)\cos(b)\cos(c) \leq \frac{1}{8}.$$ Both formulas can be derived by using elementary methods. For the first formula you only the addition formulas for cosine. Similarly, for the inequality.

Answer 3

If you are interested in an algebraic proof, mine goes as follows:

$\cos{2A} + \cos{2B} = 2 \cos{(A+B)} \cos{(A-B)} = % 2 \cos{(\pi - C)} \cos{(A-B)} = \\-2 \cos{C} \cos{(A-B)}$

Hence, \begin{eqnarray*} & \cos{2A} + \cos{2B} + \cos{2C} &\geq& - \dfrac{3}{2} \\ \iff & -2 \cos{C} \cos{(A-B)} + \cos{2C} &\geq& - \dfrac{3}{2} \\ \iff & -2 \cos{C} \cos{(A-B)} + 2\cos^2{C} - 1 &\geq& - \dfrac{3}{2} \\ \iff & 4\cos^2{C} - 4\cos{(A-B)} \cos{C} + 1 &\geq& 0 \end{eqnarray*}

We prove the last inequality with a quadratic function.

Consider $f(t) = 4t^2 - 4 \cos{\phi} \; t + 1$, $t, \phi \in \mathbb{R}$. The descriminant is given by $D = 16(\cos^2{\phi} - 1) \leq 0$. Hence $f(t) \geq 0$ for all $t, \phi \in \mathbb{R}$. The result is now obvious if we let $t = \cos{C}$ and $\phi = A - B$.

Also, $f(t) = 0$ if and only if $\phi = n\pi$, $n \in \mathbb{Z}$, which is translated to $A = B$ for triangle $ABC$. In this case, $t = 1/2$, which means that $\cos{C} = 1/2 \implies C = \frac{\pi}{3}$. Thus, equality holds if and only if $A = B = C = \frac{\pi}{3}$. i.e. if and only if the triangle is equilateral.