Mediterranean Mathematical Competition PETER O' HALLORAN MEMORIAL

We suppose that $\hat{A}\le \hat{B}\le \hat{C}$. Let $AM$ be an internal cevian of the triangle passing through $A$. Since $AM\hat{B}>\hat{C}\ge \hat{B}$, it follows that $AM>AB$. Hence it is enough to prove that: $AB<AC+AM$.

Let $\hat{C}\le 90^{\circ }$. We will prove that $c<b+h_a$, where $c=AB$, $b=AC$ and $h_a$ is the altitude from the vertex $A$. Let $H$ be the trace of the altitude from $A$. Then we have: $$AD>AB-BC\text{ and }AD>AC-DC,$$ from which we get $h_a>\frac{b+c-a}{2}$ and therefore $$b+h_a>\frac{b+c-a}{2}+b\Rightarrow \frac{c}{2}+b>\frac{c}{2}+\frac{a+b}{2}>\frac{c}{2}+\frac{c}{2}=c.$$ Since $AM\ge h_a$ we finally get $b+AM>c$.

Let $\hat{C}>90^{\circ }$. Then $b<AM$ and therefore $b+AM>b+b>b+a>c$.