$\sqrt{x^{2}+9}+\sqrt{\left(y-x\right)^{2}+4}+\sqrt{\left(9-y\right)^{2}+49}$
(x and y is not zero and different from each other.)
How can I find the least integer value of that kinda expression above ? I would like to know that which topics I lack causing me stuck and suggestions for further studying. Thanks in advance...
Let $A\equiv(0,0)$,$B\equiv(3,x)$,$C\equiv(5,y)$ and $D\equiv(12,9)$.
$\Rightarrow\sqrt{x^{2}+9}+\sqrt{\left(y-x\right)^{2}+4}+\sqrt{\left(9-y\right)^{2}+49}=AB+BC+CD\geq AD=\sqrt{12^2+9^2}=15$
Equality occurs if the points $A,B,C,D$ are collinear.
So, the least value of the given expression is $15$ which happens to be an integer.
Equation of $AD$ is :$$Y=\frac{3}{4}X$$
So, for least value $x=\frac{9}{4}$ and $y=\frac{15}{4}$