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

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.
