Is there a vector algebra/calculus proof of the fact that $\cos^{2} A + \cos^{2} B + \cos^{2} C \geq \frac{3}{4}$ for any triangle $\Delta ABC$? I have seen many algebra/trig proofs, but I was wondering if there was a way to prove this using vectors, like the way you prove the law of cosines from expanding $(\vec{u} - \vec{v})^{2}$?
This is my progress so far:
A triangle can be specified by two vectors, $\vec{u}$ and $\vec{v}$; these form two sides of a triangle (the vertex being at the origin). Then, $\vec{v} - \vec{u}$ forms the other side. Let $u = |\vec{u}|$ and $v = |\vec{v}|$, and $(\vec{v} - \vec{u})^{2} = (\vec{v} - \vec{u}) \cdot (\vec{v} - \vec{u}) = u^{2} + v^{2} - 2 \vec{u} \cdot \vec{v}$. If $A$ denotes the angle between $\vec{u}$ and $\vec{v}$ at the origin, then $\cos^{2} A = \frac{(\vec{u} \cdot \vec{v})^{2}}{u^{2} v^{2}}$. Likewise, if $B$ denotes the angle between $\vec{u}$ and $\vec{v} - \vec{u}$, then $\cos^{2} B = \frac{((\vec{v} - \vec{u}) \cdot u)^{2}}{(\vec{v} - \vec{u})^{2}u^{2}} = \frac{(\vec{v} \cdot \vec{u} - u^{2})^{2}}{(\vec{v} - \vec{u})^{2} u^{2}} = \frac{(\vec{v} \cdot \vec{u})^{2} - 2u^{2}(\vec{v}\cdot \vec{u})+u^{4}}{(\vec{v} - \vec{u})^{2}u^{2}}$. If $C$ denotes the angle between $\vec{v} - \vec{u}$ and $\vec{v}$, then $\cos^{2} C = \frac{((\vec{v} - \vec{u}) \cdot v)^{2}}{(\vec{v} - \vec{u})^{2}v^{2}} = \frac{(\vec{v} \cdot \vec{u} - v^{2})^{2}}{(\vec{v} - \vec{u})^{2} v^{2}} = \frac{(\vec{v} \cdot \vec{u})^{2} - 2v^{2}(\vec{v}\cdot \vec{u})+v^{4}}{(\vec{v} - \vec{u})^{2}v^{2}}$. Using the common denominator of $(\vec{v} - \vec{u})^{2}u^{2}v^{2}$, we then have the following: $$\cos^{2}A + \cos^{2} B + \cos^{2} C = \frac{(\vec{u} \cdot \vec{v})^{2}[(\vec{v} - \vec{u})^{2}+u^{2} + v^{2}] - 4(\vec{u} \cdot \vec{v})u^{2}v^{2} +u^{4}v^{2} + u^{2}v^{4}}{(\vec{v} - \vec{u})^{2}u^{2}v^{2}}$$ Simplifying the numerator gives: $$\cos^{2} A + \cos^{2} B + \cos^{2} C = \frac{u^{2}v^{2}(u^{2} + v^{2} - 4(\vec{u} \cdot \vec{v})) + 2(\vec{u} \cdot \vec{v})^{2}(u^{2} + v^{2} - \vec{u} \cdot \vec{v}) }{(\vec{v} - \vec{u})^{2}u^{2}v^{2}}$$
The statement that $\cos^{2} A + \cos^{2} B + \cos^{2} C \geq \frac{3}{4}$ then becomes equivalent to the statement that $$u^{2}v^{2}(u^{2}+v^{2} - 4(\vec{u} \cdot \vec{v})) + 2(\vec{u} \cdot \vec{v})(u^{2} + v^{2} - \vec{u} \cdot \vec{v}) \geq \frac{3}{4} (u^{2} + v^{2} - 2 \vec{u} \cdot \vec{v})u^{2}v^{2}$$ Subtracting the right-hand side from the left-hand side gives: $$\frac{1}{4}u^{2}v^{2}(u^{2} + v^{2} - 10(\vec{u} \cdot \vec{v})) + 2(\vec{u} \cdot \vec{v})(u^{2} + v^{2} - \vec{u} \cdot \vec{v}) \geq 0$$ One thought I had to prove this inequality was to rewrite the left-hand side as follows:
$$\frac{1}{4}v^{2}(u^{2} - 2(\vec{u} \cdot \vec{v}))^{2} + \frac{1}{4}u^{2}(v^{2} - 2(\vec{u} \cdot \vec{v}))^{2} + (\vec{u} \cdot \vec{v})^{2} - \frac{1}{2}(\vec{u} \cdot \vec{v})u^{2}v^{2} - (\vec{u} \cdot \vec{v})^{3}$$
It would then suffice to show that $ (\vec{u} \cdot \vec{v})^{2} - \frac{1}{2}(\vec{u} \cdot \vec{v})u^{2}v^{2} - (\vec{u} \cdot \vec{v})^{3} \geq 0$ or equivalently that $(\vec{u} \cdot \vec{v}) ((\vec{u} \cdot \vec{v})-\frac{1}{2}u^{2}v^{2} -(\vec{u} \cdot \vec{v})^{2}) \geq 0$. However, if $\vec{u} \cdot \vec{v} = \frac{\sqrt{3}}{2}$ and $u^{2} = v^{2} = 1$ (angle $A$ is $30^{\circ}$), then $(\vec{u} \cdot \vec{v})(\vec{u} \cdot \vec{v} - \frac{1}{2}u^{2}v^{2}-(\vec{u} \cdot \vec{v})) = \frac{1}{8}(6- 5\sqrt{3}) < 0$.
Is there another way to rewrite $$\frac{1}{4}u^{2}v^{2}(u^{2} + v^{2} - 10(\vec{u} \cdot \vec{v})) + 2(\vec{u} \cdot \vec{v})(u^{2} + v^{2} - \vec{u} \cdot \vec{v})$$ as a linear combination of $(u^{2} - 2\vec{u} \cdot \vec{v})^{2}$ and $(v^{2} - 2\vec{u} \cdot \vec{v})^{2}$ (with coefficients that are all nonnegative?) Note that the equality case should imply that $A = B = C = 60^{\circ}$, which should be equivalent, in this context, to $u^{2} = 2 \vec{u} \cdot \vec{v}$ and $v^{2} = 2\vec{u} \cdot \vec{v}$.
$A = 60^{\circ} \iff u^{2}v^{2}=4(u \cdot v)^{2}$;
$B = 60^{\circ} \iff (\vec{u} - \vec{v})^{2}u^{2} = 4(\vec{u} \cdot (\vec{v} - \vec{u}))^{2} \iff u^{4} + u^{2}v^{2} - 2(\vec{u} \cdot \vec{v})u^{2} = 4u^{4} + 4(\vec{u} \cdot \vec{v})^{2}- 8(\vec{u} \cdot \vec{v})u^{2} \iff 3u^{4} - 6(\vec{u} \cdot \vec{v})u^{2} = 0 \iff u^{2} = 2 u \cdot v$$
$C= 60^{\circ} \iff (\vec{u} - \vec{v})^{2}v^{2} = 4(\vec{u} \cdot (\vec{v} - \vec{u}))^{2} \iff v^{4} + u^{2}v^{2} - 2(\vec{u} \cdot \vec{v})v^{2} = 4v^{4} + 4(\vec{u} \cdot \vec{v})^{2}- 8(\vec{u} \cdot \vec{v})v^{2} \iff 3v^{4} - 6(\vec{u} \cdot \vec{v})v^{2} = 0 \iff v^{2} = 2 \vec{u} \cdot \vec{v}$ )
To reduce the difficulty of multivariable calculus, we can express $C$ in terms of $A, B$, since $A + B + C = \pi$
Denote $f(A, B) := \cos^2{A} + \cos^2{B} + \cos^2{(A + B) = k}$, where $k$ is a constant.
Now, WLOG, take a partial derivative w.r.t, $A$. $$\begin{align}\\& \dfrac{d[f(A, B)]}{dA} := \\& -\sin{2A} \\& - \dfrac{dB}{dA}\sin{2B} \\& -2\sin{(A + B)}\cos{(A + B)} -2\dfrac{dB}{dA}\sin{(A + B)}\cos{(A + B)} \\& = 0\end{align}$$
$$\text{Since, } \dfrac{dB}{dA} = 0 \\ \to -\sin{2A} -2\sin{(A + B)}\cos{(A + B)} = 0 \\ \to -\sin{2A} = \sin{(2A + 2B)} \\ \to -\sin{2A} = \sin{2A}\cdot\cos{2B} + \cos{2A} \cdot \sin{2B} \\ \to -1 = \cos{2B} + \cot{2A} \cdot \sin{2B} \\ \to -\csc{2B} = \cot{2B} + \cot{2A}$$ $$\\ \to -\dfrac{1 + \cos{2B}}{\sin{2B}} = \cot{2A}$$
$$\therefore -\cot^{-1}{\left( \cot{B} \right)} = 2A$$ $$\\ \therefore A = \dfrac{n\pi - B}{2}$$
$$\\ \therefore \cos^2{\dfrac{n\pi - B}{2}} + \cos^2{B} + \cos^2{ \dfrac{n\pi + B}{2}} \\= 2\sin^2{\dfrac{B}{2}} + \cos^2{B}$$.
Now, we have our equation in terms of $1$ variable. Hence, redo the differentiation w.r.t to $B$ to find our minimum:
$$\dfrac{d}{dB}\left[ 2\sin^2{\dfrac{B}{2}} + \cos^2{B} \right] = 0 \\ \to 2\sin{\dfrac{B}{2}}\cos{\dfrac{B}{2}} - 2\sin{B}\cos{B} = 0 \\ \to \sin{B} = \sin{2B} \\ \to B = \dfrac{(6n \pm 1)\pi}{3}, n\pi \\ \text{However, }0 < B < \pi, \text{ and } \therefore B = \dfrac{\pi}{3}$$
This is confirmed through $2\sin^2{\dfrac{\pi/3}{2}} + \cos^2{\dfrac{\pi}{3}} = \dfrac{3}{4}$.
We can describe it is a minimum through the second derivative test;
$\dfrac{d^2}{dB^2} = \dfrac{3}{2} > 0 \text{ when } B = \dfrac{\pi}{3}$ and it is the only minimum within the domain of $0 < B < \pi$