Minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12$

$\sqrt{x^2+25}$ is the distance between $(x,0)$ and $(0,5)$, while $\sqrt{(x-12)^2+16}$ is the distance between $(x,0)$ and $(12,-4)$. Hence we need to find $x$ such that the sum of these two distances reaches the minimum. That is exactly the point that the line through $(12,-4)$ and $(0,5)$ meets the $x$-axis. Therefore we get $x = \frac{20}{3}$ and so $y = 12-\frac{20}{3} = \frac{16}{3}$.

We are asked for the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12;$

that is, the minimum value of $\sqrt{x^2+25}+\sqrt{(12-x)^2+16}.$

Let $\overrightarrow a=(x,5)$ and $\overrightarrow b=(12-x,4)$ in $\mathbb R^2,$ so $\overrightarrow a + \overrightarrow b=(12,9).$

By the triangle inequality, $|\overrightarrow a+\overrightarrow b|\le|\overrightarrow a|+|\overrightarrow b|.$

Therefore, $\mathbf{15}=\sqrt{12^2+9^2}\le\sqrt{x^2+25}+\sqrt{(12-x)^2+16}.$

Here's an algebraic solution.

Let $x = z + 20/3$ then $$ \sqrt{x^2+25}+\sqrt{y^2+16}=\sqrt{x^2+25}+\sqrt{(x-12)^2+16}\\ =\sqrt{(z+20/3)^2+25}+\sqrt{(z-16/3)^2+16}\\ = \sqrt{z^2 + 40 z/3 + 625/9}+\sqrt{z^2 - 32 z/3 + 400/9} = f(z) $$ Then $$ f(z)^2= 1025/9 + 2 z^2 + 8z/3 + 2 \sqrt{z^2 + 40 z/3 + 625/9}\sqrt{z^2 - 32 z/3 + 400/9} \\ = 1025/9 + 2 z^2 + 8z/3 + 2 \sqrt{z^4 + (8 z^3)/3 - (85 z^2)/3 - (4000 z)/27 + 250000/81} $$ Since $f(z)$ is always positive, we can claim that $f(z)^2 \ge 15^2 = 225$ so we need to show that $$ 1025/9 + 2 z^2 + 8z/3 + 2 \sqrt{z^2 + 40 z/3 + 625/9}\sqrt{z^2 - 32 z/3 + 400/9} \ge 2025/9\\ \leftrightarrow \sqrt{z^4 + (8 z^3)/3 - (85 z^2)/3 - (4000 z)/27 + 250000/81} \ge 500/9 - z^2 - 4z/3 \\ \leftrightarrow z^4 + (8 z^3)/3 - (85 z^2)/3 - (4000 z)/27 + 250000/81 \ge (500/9 - z^2 - 4z/3)^2 = \\z^4 + (8 z^3)/3 - (328 z^2)/3 - (4000 z)/27 + 250000/81 \\ \leftrightarrow - (85 z^2)/3 \ge - (328 z^2)/3 $$ and this is obviously true, with equality for $z=0$.