Length of median is less than half of adjacent sides

[Mr Davis 97]

Homework Statement

$ABC$ is a triangle, the midpoint of $AB$ is $H$. Prove that $2CH < AC+CB$.

Homework Equations

The Attempt at a Solution

Note that by the triangle inequality that $CH \le HA + AC$ and that $CH \le HB + BC$. Adding these two inequalities gives $$2CH \le HA+HB+AC+CB = AB+AC+CB < AC+CB.$$

This problem is from a problem-solving book, but it seems way too easy and uninteresting. Am I making some egregious error, or is it just easy?

 

[Mark44]

Homework Statement

$ABC$ is a triangle, the midpoint of $AB$ is $H$. Prove that $2CH < AC+CB$.

Homework Equations

The Attempt at a Solution

Note that by the triangle inequality that $CH \le HA + AC$ and that $CH \le HB + BC$. Adding these two inequalities gives $$2CH \le HA+HB+AC+CB = AB+AC+CB < AC+CB.$$

This problem is from a problem-solving book, but it seems way too easy and uninteresting. Am I making some egregious error, or is it just easy?

I think it is easy. Draw a picture of a triangle, labelling the vertices and the midpoint H. Now form a parallelogram ABCD, with |DA| = |BC| and |DB| = |AC|. Extend CH to point D, so that it is the long diagonal of the parallelogram. Clearly this diagonal is shorter than the sum of the two sides AC and CB of the triangle.