$\sqrt {A^2+(N*A)^2}+\sqrt {B^2+(N*B)^2}<\sqrt {A^2+(N*A+M)^2}+\sqrt {B^2+(N*B-M)^2}$
Where $A$ and $B$ are positive real numbers, and $N$ and $M$ are real numbers, also $M\neq0$
I and J are segments that end in a single point on the lowest segment, and start at the top of A and B
$I+J$ is equal to the left side of the inequation, and has the lowest value possible given the length of the lowest segment, the length of $A$ and the length of $B$, i want to prove that $I+J$ is minimal
The inequality is valid if and only if the angles formed at the horizontal line segment are equal, which, in fact, they are. You can prove the inequality by reflecting the second segment across the horizontal and observing that the path from the upper vertex to the (reflected) lower vertex is shortest when you have the straight line segment joining them. This happens precisely when those angles are equal.
COMMENT: If you labeled the upper left vertex $E$, the right vertex $F$, the point on the horizontal $G$, and the reflected right vertex $F'$, then my statement says that $EG + GF'$ is smallest when $G$ lies on the line segment $\overline{EF'}$.