Prove that in any triangle four times the sum of the medians is greater than three times the perimeter

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Please find below the solution to the asked query:

We form our diagram , As :

Here , M_a , M_b  and M_c  are medians on sides BC , AC and AB respectively . G is centroid

We know centroid divide median in ratio of 2 : 1 

In $∆$ ABG we know from triangle inequality theorem :

AG  + BG  >  AB , So

$\frac{2}{3}$M_a  + $\frac{2}{3}$ M_b > AB                                                             --- ( 1 )

And in $∆$ BCG we know from triangle inequality theorem :

BG  + CG  >  AB , So

$\frac{2}{3}$M_b  + $\frac{2}{3}$ M_c > BC                                                             --- ( 2 )

And in $∆$ ACG we know from triangle inequality theorem :

AG  + CG  >  AC , So

$\frac{2}{3}$M_a  + $\frac{2}{3}$ M_c > AB                                                             --- ( 3 )

Now we add equation 1 , 2 and 3 and get


$\frac{4}{3}$M_a  + $\frac{4}{3}$ M_b + $\frac{4}{3}$ M_c > AB + BC  +  AC

$\frac{4}{3}$( M_a  +  M_b +  M_c ) > AB + BC  +  AC

4 ( M_a  +  M_b +  M_c ) > 3 ( AB + BC  +  AC )

Therefore,

4 ( Sum of medians ) >  3 ( Perimeter )                                                ( Hence proved )


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