Foot of the altitude $K$ divides side $c$ into ratio $1:2$, prove $3(a-b) < c$

Question

Foot of the altitude $K$ from vertex $C$ in triangle $ABC$ divides side $c$ in ratio $1:2$, proof inequality $3(a-b)<c$ if you know, that $a>b$.

Triangle sketch:

Answer 1

By Pythagorean's theorem: $$a^2-h^2=(2x)^2=\frac{4c^2}{9},\quad b^2-h^2=x^2=\frac{c^2}{9}.$$ Hence $$a^2-b^2=\frac{c^2}{3}\quad\Rightarrow 3(a-b)=c\cdot\frac{c}{a+b}<c,$$ where the triangle inequality $a+b>c$ plays the essential role.