x=2，y=1.5

triangle inequality $\sqrt{x^2+y^2}+\sqrt{(x-4)^2+(y-3)^2} \geq \sqrt{(4-x+x)^2+(y+3-y)^2}= 5$ equality only for $\frac{x}{4-x} = \frac{y}{3-y}$ $3x-xy=4y-xy$ $3x=4y$ $8xy = 24$ $xy=3$ $y^2 = \frac{9}{4}$ $x,y = (\frac{-3}{2},-2)$ doesnt satisfy $x,y = (2,\frac{3}{2})$

StarLex1 wrote: triangle inequality $\sqrt{x^2+y^2}+\sqrt{(x-4)^2+(y-3)^2} \geq \sqrt{(4-x+x)^2+(y+3-y)^2}= 5$ equality only for $\frac{x}{4-x} = \frac{y}{3-y}$ $3x-xy=4y-xy$ $3x=4y$ $8xy = 24$ $xy=3$ $y^2 = \frac{9}{4}$ $x,y = (\frac{-3}{2},-2)$ doesnt satisfy $x,y = (2,\frac{3}{2})$ Yeah a solution with minkowki is very good,but there is a decent solution like taking the first expression to the right and squaring both sides,this will lead a very good position

Just take the first expression to the RHS and square both side!

$\sqrt{x^2+y^2}+\sqrt{(x-4)^2+(y-3)^2}=5$, this represents that the sum of the distances from a point $(x,y)$ to $(0,0)$ and $(3,4)$, the distance between 3,4 and origin is $5$ so the point lies on $y=\frac{3x}{4}$, subsituting into second equation, $(x,y)=(2, 1.5)$ (note that the x coord needs to be positive)

$$\sqrt{x^2+y^2}\hspace{1mm}=\hspace{1mm}5\hspace{1mm}-\hspace{1mm}\sqrt{(x-4)^2+(y-3^2)}\Rightarrow x^2\hspace{1mm}+\hspace{1mm}y^2\hspace{1mm}=\hspace{1mm}25\hspace{1mm}+\hspace{1mm}x^2\hspace{1mm}-\hspace{1mm}8x\hspace{1mm}+\hspace{1mm}16\hspace{1mm}+\hspace{1mm}y^2\hspace{1mm}-\hspace{1mm}6y\hspace{1mm}+\hspace{1mm}9\hspace{1mm}-\hspace{1mm}10\sqrt{(x-4)^2+(y-3)^2}$$ $$50\hspace{1mm}-\hspace{1mm}8x\hspace{1mm}-\hspace{1mm}6y\hspace{1mm}=\hspace{1mm}10\sqrt{(x-4)^2+(y-3)^2}$$ Divide both side by 2 and square both sides $$625\hspace{1mm}+\hspace{1mm}16x^2\hspace{1mm}+\hspace{1mm}9y^2\hspace{1mm}-\hspace{1mm}200x\hspace{1mm}-\hspace{1mm}150y\hspace{1mm}+\hspace{1mm}24xy\hspace{1mm}=\hspace{1mm}25x^2\hspace{1mm}-\hspace{1mm}200x\hspace{1mm}\hspace{1mm}+\hspace{1mm}25y^2\hspace{1mm}-\hspace{1mm}150y\hspace{1mm}+\hspace{1mm}625$$ $$9x^2\hspace{1mm}+\hspace{1mm}16y^2\hspace{1mm}=\hspace{1mm}24xy\Rightarrow (3x-4y)^2=0\Rightarrow 3x=4y$$ $$3x^2\hspace{1mm}+\hspace{1mm}4xy\hspace{1mm}=\hspace{1mm}3x^2\hspace{1mm}+\hspace{1mm}3x^2\hspace{1mm}=\hspace{1mm}6x^2\hspace{1mm}=\hspace{1mm}24\Rightarrow x=2\hspace{3mm}y=1,5$$

It's just Minkowski's Inequality. The rest follows directly.

Minkowski theorem kills this problem.

Good solution with Minkovski