Triangle inequalities in proof If we called the perimeter of some triangle $b$. Prove that if you added the lengths of any two of its medians
(i) It would not be not larger than $\frac{3b}{4}$
(ii) It would not be smaller than $\frac{3b}{8}$
This came up in a math competition that I did in October and I think that using the triangle inequality will be necessary. However, I only know how to prove it when we add all of the medians.
What should I do? 
Thank you.
 A: (ii) should be fixed. It would be larger than $\frac{3p}{8}$.
Let $AD, BE, CF$ be the medians, and let the centroid be $G$.
WLOG, let the two medians be $m_a, m_b$.
From the Triangle Inequality on $\triangle AGB$, we have $$\frac{2}{3}m_a+\frac{2}{3}m_b > c$$
Summing this cyclically gives $$m_a+m_b+m_c > \frac{3}{4}(a+b+c)$$
Let the midpoint of $GB$ be $M$. We have $$GD=\frac{1}{3}m_a, GM=\frac{1}{3}m_b, MD=\frac{1}{3}m_c$$
Therefore, triangle inequality on $\triangle GMD$ gives $m_a+m_b>m_c$.
Therefore, we have $2(m_a+m_b) > \frac{3}{4}(a+b+c)$, so $m_a+m_b>\frac{3}{8}(a+b+c)$.
A: As the second part is proven in the other answer, I will add the proof for the first part here. I will use the same point system as described in the other answer.
WLOG we consider the sum $m_a+m_b$.
Let the midpoint of $CF$ be $R$. Connect $BR.AR$.
First we show $BR+AR>BG+AG$. This is true because $G$ is inside triangle $BRA$. If we let the intersection of $AG$ and $BR$ be $T$, then $BG+AG<BT+AT<BR+AR$.
Now Connect $RD$. Since $BD+RD>BR$ we know $BC+BF>2BR$ since $BC=2BD$ and $BF=2RD$.
Similarly we have $AC+AF>2AR$.
Sum the two up we get $AB+BC+CA>2AR+2BR>2AG+2BG=2\cdot{2\over3}(m_a+m_b)$ and the result follows.
