Minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12$ If $x,y\in\mathbb R^+$, $x+y=12$, what is the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$?
I got the question from a mathematical olympiad competition (of China) for secondary 2 student, so I don't expect an "analysis" answer.
The answer should be $15$, when $x=\frac{20}{3}$ and $ y=\frac{16}{3}$ (just answer, no solution :< ).
I try to use AM-GM inequality, but I couldn't manage to get the answer.
$$\sqrt{x^2+25}+\sqrt{y^2+16}=\sqrt{x^2+25}+\sqrt{(x-12)^2+16}=\sqrt{x^2+25}+\sqrt{x^2-24x+160}$$
Can this help?
I also tried to plot the graph, see here.
Any help would be appreciated. Thx!

Let 
$f = \sqrt{x^2 + 25} + \sqrt{y^2 + 16}$
$g = x+y -12 =0 $
By Lagrange's Undetermined multipliers method,
$F = f + g\lambda = \sqrt{x^2 + 25} + \sqrt{y^2 + 16} + \lambda(x+y -12)$
Differentiating partially w.r.t x and y and using $F_x = 0$ and $F_y = 0 $ at extremum,
$F_x = \frac{x}{\sqrt{x^2 + 25}} +\lambda = 0$
$F_y = \frac{y}{\sqrt{y^2 + 16}} +\lambda = 0$
So, 
$\frac{x}{\sqrt{x^2 + 25}} = \frac{y}{\sqrt{y^2 + 16}}$
$\frac{x^2}{x^2 + 25} = \frac{y^2}{y^2 + 16}$
$\frac{x^2 + 25}{x^2} = \frac{y^2 + 16}{y^2}$
$1 + \frac{25}{x^2} = 1 +  \frac{16}{y^2}$
On solving,
$5y = \pm 4x$
Using this in g,
$x \pm 4x/5 = 12$
$x = 20/3, 60$
Using these values, 
$y = 12 - x $
$y = 12 -20/3 = 16/3$ and $y =  12 - 60 = -48$
Using these values of x and y in f we find its value is minimum at $(\frac{20}{3},\frac{16}{3})$.
