<s>[INST] Minimize $3\sqrt{5-2x}+\sqrt{13-6y}$ subject to $x^2+y^2=4$ 
If $x, y \in \mathbb{R}$ such that $x^2+y^2=4$, find the minimum value of $3\sqrt{5-2x}+\sqrt{13-6y}$.

I could observe that we can write
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{x^2+y^2+1-2x}+\sqrt{x^2+y^2+9-6y}$$
$\implies$
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-3)^2}=3PA+PB$$
Where $P$ is a generic point on $x^2+y^2=4$ and $A(1,0),B(0,3)$.
So the problem essentially means which point on the circle $x^2+y^2=4$ minimizes $3PA+PB$.
I am really struggling to find the geometrical notion of this and hence unable to solve.
 [/INST] Comment: you may use this modification:
We rewrite final relation as:
$$A=\sqrt{[3(x-1)=a]^2+(3y=b)^2}+\sqrt{(x=a')^2+(y-3=b')^2}$$
and use this inequality:
$\sqrt{a^2+b^2}+\sqrt{a'^2+b'^2}\geq\sqrt{(a+a')^2+(b+b'^2)}$
we get:
$A\geq\sqrt{16(x^2+y^2)-24(x+y)+18}$
$x^2+y^2=4$
$\Rightarrow$
$A\geq \sqrt{82-24(x+y)}$
If $x=y=\sqrt 2$ then $A\geq\sqrt{84-67.7}\approx 4$
Update:Wolfram says minimum is $2\sqrt {10}=6.32$ at $(x,y)=(\frac25+\frac{3\sqrt6}5=1.87, \frac 65-\frac{\sqrt6}5)=0.71$. If we put this in $\sqrt{82-24(x+y)}$ we get 4.47. mind you x and y must suffice $x^2+y^2=4$, and what Wolfram gives does; $1.87^2+0.71^2=4$.Hence 40 can not be minimum .
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