<s>[INST] Compare the sum of the squares of the median of a triangle to the sum of the squares of sides You have to compare the sum of the squares of the median of a triangle to the sum of the squares of sides?
 [/INST] 
$$ 3( AB^2 + BC^2 + AC^2 ) = 4 ( AD^2 + BE^2 + CF^2)$$

AB , BC , CD are lengths of sides of triangle and AD , BE , CF are lengths of medians of triangle
HINT : 
start with appollonius theorem and add all 
Apollonius Theorem
so you will get relation as 
3 (sum of square of sides) = 4 ( sum of squares of medians )
Proof : 

$$ AB^2 + AC^2 = 2 ( AD^2 + \frac{BC^2}{4} )$$
$$ AB^2 + BC^2 = 2 ( BE^2 + \frac{AC^2}{4} )$$
$$ AC^2 + BC^2 = 2 ( CF^2 + \frac{AB^2}{4} )$$
Add all above and bring sides to LHS and medians to RHS to get
$$ 3( AB^2 + BC^2 + AC^2 ) = 4 ( AD^2 + BE^2 + CF^2)$$
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