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, $f(x)_{min}=\sqrt{15}$.

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 $(\cos{x},\sin{x})=\left(-\frac{3}{5},-\frac{4}{5}\right).$
