# Prove, in a $\Delta ABC$ with medians $BE, CF$, $BE + CF > BC\cdot\frac32$

Consider a $\Delta ABC$ with medians $BE$ and $CF$. Prove that

$$BE + CF > BC\cdot\frac32$$

Consider the following inequalities, given by the triangle inequality:

$$BE + CE > BC \implies BE + \frac{AC}{2} > BC$$

$$\implies BE > BC - \frac{AC}{2} \tag1$$

Similarly,

$$CF > BC - \frac{AB}{2} \tag2$$

Adding the $(1)$ and $(2)$ gives us:

$$BE + CF > 2BC - \left(\frac{AB + AC}{2}\right) \tag3$$

Now,

$$AB + AC > BC$$

$$\implies -\frac{AB+AC}{2} < -\frac{BC}{2}$$

So, replacing $\frac{AB + AC}{2}$ by $\frac{BC}{2}$ in $(3)$ won't work. How do I proceed?

See in $\bigtriangleup BGC$ where $G$ is centroid:

• $BG+GC >BC$
• Now $BG=\frac{2}{3}BE$ and $GC=\frac{2}{3}CF$.
• So putting the values we will get $BE+FC>\frac{3}{2}BC$.

Join $EF=\frac12BC$

Let $G$ be the intersection of $BE,CF$

Now in $\triangle BGC, BG+CG>BC$ and in $\triangle EFG, GF+GE>EF=\frac12BC$

Now add them and use $BG+GE=BE$ and $CG+GF=CF$