Find the minimum of $\sqrt{\cos x+3}+\sqrt{2\sin x+7}$ without derivative How do we find the minimum of
$$f(x)=\sqrt{\cos x+3}+\sqrt{2\sin x+7}$$
without using derivatives?
This problem is probably related to circles of Apollonius.
I have tried AM-GM and Cauchy-Schwarz inequality but I can't work it out.
Anyway, I have solved it in a more geometric way. Here's my answer.
Firstly we can do some identical transformation.
$$f(x)=\dfrac{\sqrt{2}}{2}(\sqrt{(\cos x+1)^2+(\sin x)^2+4}+\sqrt{(\cos x)^2+(\sin x+2)^2+9})$$
So that it makes sense in geometry.

$P(\cos x,\sin x)$ is on the circle $x^2+y^2=1$, and the value of $f(x)$ equals to sum of the distance from $A(0,-2)$ to $P$ and from $B(-1,0)$ to $P$.
In other words:
$$f(x)=\dfrac{\sqrt{2}}{2}(\sqrt{|PB|^2+4}+\sqrt{|PA|^2+9}).$$
And here we can use Minkowski inequality.
$$f(x)\geq \dfrac{\sqrt{2}}{2} \sqrt{(|PA|+|PB|)^2+25}$$
When $P$,$A$,$B$ is collinear, $RHS$ gets the minimum. Meanwhile, $LHS = RHS$.
Therefore, $f(x)_{min}=\sqrt{15}$.
 A: Here's a rough idea for a solution (since there are some details I skim over). I'm pretty sure the algebra can be simplified using some AM-GM variation, but I couldn't make it work. Although I'm hopeful that someone else can give a more optimal solution.
Note that for all $t \in \mathbb{R}$
\begin{align*}
&0 \le(t+2)^2(5t^2+6) =5 t^4 + 20 t^3 + 26 t^2 + 24 t + 24\\
\color{darkblue}{\implies}& 14 t^4 + 8 t^3 + 42 t^2 + 16 t + 28 \ge 9t^4 - 12 t^3 + 16 t^2 - 8 t + 4\\
\color{darkblue}{\implies}&\left(2t^2 +4\right)\left(
4t +7t^2+7\right)\ge (3 t^2 + 2 - 2 t)^2\\
\color{darkblue}{\implies}&2\sqrt{2t^2 +4}\sqrt{4t+7t^2+7} \ge 2\left(3t^2 +2-2t \right)\\
\color{darkblue}{\implies}&2t^2 +4 +2\sqrt{2t^2 +4}\sqrt{4t+7t^2+7} + 4t+7t^2 +7 \ge 15\left(t^2+1\right)\\
\color{darkblue}{\implies}&\left(\sqrt{2t^2 +4} +  \sqrt{4t+7t^2+7}\right)^2 \ge\left( \sqrt{15}\sqrt{t^2+1}\right)^2\\
\color{darkblue}{\implies}&\frac{\sqrt{2t^2+4}}{\sqrt{t^2+1}} + \frac{\sqrt{4t+7t^2+7}}{\sqrt{t^2+1}}\ge \sqrt{15}\\
\color{darkblue}{\implies}&\sqrt{\frac{1-t^{2}}{1+t^{2}}+3}\ +\ \sqrt{2\frac{2t}{1+t^{2}}+7} \ge \sqrt{15}
\end{align*}
Lastly, notice that under the half-angle substitution the problem becomes showing that
$$
\min\left\{\sqrt{\frac{1-t^{2}}{1+t^{2}}+3}\ +\ \sqrt{2\frac{2t}{1+t^{2}}+7}\right\} = \sqrt{15}
$$
where $t = \tan(x/2)$. Since $\sqrt{\frac{1-t^{2}}{1+t^{2}}+3}\ +\ \sqrt{\frac{4t}{1+t^{2}}+7}\Bigg\vert_{t=-2} = \sqrt{15}$ we are done.
A: Because $$\sqrt{\cos x+3}+\sqrt{2\sin x+7}=$$
$$=\sqrt{15}+\left(\sqrt{\cos x+3}-2\sqrt{\frac{3}{5}}+\frac{5\sqrt5}{24\sqrt3}\left(\cos^2x-\frac{9}{25}\right)\right)+$$
$$+\left(\sqrt{2\sin x+7}-3\sqrt{\frac{3}{5}}+\frac{5\sqrt5}{24\sqrt3}\left(\sin^2x-\frac{16}{25}\right)\right)\geq\sqrt{15}.$$
The equality occurs for $(\cos{x},\sin{x})=\left(-\frac{3}{5},-\frac{4}{5}\right).$
A: This is not a proof but a sequence of heuristic and "nicety" arguments which lead to $\sqrt{15}$.
Let $f(x)$ be the expression to be minimized.
The substitution $\sin(x)\to t$, $\cos(x) \to \pm \sqrt{1-t^2}$ gives
$$f(x(t)) = \sqrt{3-\sqrt{1-t^2}}+\sqrt{2 t+7}$$
where we have taken the minus sign under the square root because this leads to a smaller expression.
Now letting $t\to \frac {p}{q}$ gives
$$f=\frac{1}{\sqrt{q}}
\left(\sqrt{3 q-\sqrt{q^2-p^2}}+\sqrt{2 p+7 q}\right)$$
The second term in the bracket suggests a negative value for $p$ ($q$ can't be negative because of the common factor in front) and the square root term under square root in the first term suggests that $p$ and $q$ might be members of a Pythagorean triple, like e.g. $(3,4,5)$.  This leads us to test $(p,q) = (-4,5)$ or $t=-\frac{4}{5}$ which gives $f=\sqrt {15}$. This is the minumum value of $f$ as other users have shown.
