The sum of two vectors $\mathbf{A}$ and $\mathbf{B}$ is $\mathbf{C}$. Magnitude of vector $\mathbf{A}$ is twice that of vector $\mathbf{B}$. Then show that $\mathbf{C}$ cannot make an angle greater than $30^\circ$ with vector $\mathbf{A}$, without using differential calculus.
I tried to construct a perpendicular in the figure. But at least I am coming to the conclusion that the $\cos$ function on the angle between two vectors is less than $1/3$
On
$$\newcommand{\tx}[1]{\mathrm{#1}}\newcommand{\bx}[1]{\mathbf{#1}}$$Using boldface $\bx{A},\bx{B},\bx{C}$ to denote vectors and $\tx{A},\tx{B},\tx{C}$ to denote their magnitude we have:
$$\bx{A} + \bx{B} = \bx{C}\tag{1}$$ $$\tx{A} = 2\tx{B}\tag{2}$$ Let's put $\alpha = \angle({\bx{A},\bx{C}})$ (the angle between $\bx{A}$ and $\bx{C}$) and likewise $\beta = \angle({\bx{A},\bx{B}})$.
So note that:
\begin{align*} \tx{C}^2 &= \bx{C}\cdot\bx{C} \\ &= (\bx{A}+\bx{B})\cdot(\bx{A}+\bx{B}) \\ &= \tx{A}^2+\tx{B}^2+2\tx{A}\tx{B}\cos\beta\\ &= 5\tx{B}^2+4\tx{B}^2\cos\beta\tag{3} \end{align*}
Now use the definition of the dot product and apply the relations $\text{(1),(2),(3)}$:
\begin{align*} \cos\alpha &= \frac{\bx{A}\cdot\bx{C}}{\tx{A}\tx{C}} \\ &= \frac{\bx{A}\cdot(\bx{A}+\bx{B})}{\tx{A}\sqrt{5\tx{B}^2+4\tx{B}^2\cos\beta}} \\ &= \frac{\tx{A}^2+\tx{A}\tx{B}\cos\beta}{\tx{A}\sqrt{5\tx{B}^2+4\tx{B}^2\cos\beta}} \\ &= \frac{\tx{A}+\tx{B}\cos\beta}{\sqrt{5\tx{B}^2+4\tx{B}^2\cos\beta}} \\ &= \frac{2\tx{B}+\tx{B}\cos\beta}{\sqrt{5\tx{B}^2+4\tx{B}^2\cos\beta}} \\ &= \frac{2+\cos\beta}{\sqrt{5+4\cos\beta}} \end{align*}
Now, we want to show that angle $\alpha$ can't exceed $30^\circ$ so let's assume by contradiction that:
$$\alpha \gt \frac{\pi}{6} \implies \cos\alpha \lt \frac{\sqrt{3}}{2} $$
So we have:
$$ \frac{2+\cos\beta}{\sqrt{5+4\cos\beta}} < \frac{\sqrt{3}}{2} $$
Since it's clear that the LHS's numerator is always positive (minimized at $2-1=1$) then we can safely square both sides:
\begin{aligned} \frac{4+\cos^2\beta+4\cos\beta}{5+4\cos\beta} < \frac{3}{4} &\implies 16+4\cos^2\beta+16\cos\beta<15+12\cos\beta \\ &\implies 4\cos^2\beta+4\cos\beta+1 < 0 \\ &\implies (2\cos\beta+1)^2 < 0 \end{aligned}
Which is clearly never satisfied for any $\beta$. Therefore by contradiction we have shown that $\alpha \leq \frac{\pi}{6}$ $\Box$
By at most two rotations, you can get the two vectors to be in the x-y plane, $\vec A=\hat{\vec x}$ and then $\vec B=-\hat{\vec x}\frac12\cos\varphi+\hat{\vec y}\frac12\sin\varphi$ so that $\vec C=\hat{\vec x}(1-\frac12\cos\varphi)+\hat{\vec y}\frac12\sin\varphi$
Using the cross product $$\vec A\times\vec C=\hat{\vec z}\frac12\sin\varphi =|\vec A||\vec C|\sin\vartheta\ \hat{\vec n}\\ \sin\vartheta=\frac{\frac12\sin\varphi}{(1)\sqrt{(1-\frac12\cos\varphi)^2+\frac14\sin^2\varphi}}\\ =\frac{\sin\varphi}{\sqrt{5-4\cos\varphi}}$$ Now we consider what are the possible ranges we need to inspect. We start with $\varphi\in(-\pi,+\pi]$. We do not have to check $\varphi\leqslant0$ because the angle subtended will just be equal in magnitude to the case where $\varphi\geqslant0$, and we also do not need to inspect $[\frac\pi2,\pi]$ because numerator is monotonically decreasing and denominator is monotonically increasing on that interval. Hence, we only have to inspect $\varphi\in(0,\pi/2)$. We just note here that $\varphi=0\implies\sin\vartheta=0\bigwedge\varphi=\pi/2\implies\sin\vartheta=1/\sqrt5$. Next, we apply the double angle formulæ and square everything, $$\sin^2\vartheta (1+8\sin^2\frac\varphi2)=4\sin^2\frac\varphi2\cos^2\frac\varphi2\\ \text{Let }\chi=\sin^2\frac\varphi2\qquad\implies\qquad\sin^2\vartheta(1+8\chi)=4\chi(1-\chi)\\ (2\chi)^2+2(2\chi)(2\sin^2\vartheta-1)+\sin^2\vartheta=0$$ In this last equation, for most values of $\sin^2\vartheta$, there will be two roots. But for the stationary value of $\sin^2\vartheta,$ there will only be one repeated root. That is the case of the $0=$ Discriminant $=b^2-4ac$ $$(2\sin^2\vartheta-1)^2-\sin^2\vartheta=0\\ (2\sin^2\vartheta-1-\sin\vartheta)(2\sin^2\vartheta-1+\sin\vartheta)=0\\ (2\sin\vartheta+1)(\sin\vartheta-1)(2\sin\vartheta-1)(\sin\vartheta+1)=0\\ \sin\vartheta=-1,-\frac12,+\frac12,+1$$ We can immediately ignore the negative values because $\varphi\in(0,\pi/2)\implies\sin\vartheta>0$, and then we can reject $\sin\vartheta=+1$ because that gives a negative $\chi=\sin^2\frac\varphi2$, which is impossible in the real numbers. The only acceptable value is thus $\sin\vartheta_\text{max}=\dfrac12$ which, working backwards, comes from $\varphi=\dfrac\pi3$. This value is clearly the maximum because $0<\dfrac1{\sqrt5}<\dfrac12$ $$\sin\vartheta_\text{max}=\frac12\qquad\implies\qquad\vartheta_\text{max}=\frac\pi6=30^\circ$$ QED