If G is the centroid of ΔABC such that $\vec{GB}$ and $\vec{GC}$ are inclined at on obtuse angle, then

$5a^2 >b^2 + c^2$

Taking A as the origin, let the position vectors of B and C be $\vec b$ and $\vec c$ respectively. Then, the position vector of G is $\frac{\vec b+\vec c}{3}$.

Since $\vec{GB}$ and $\vec{GC}$ are inclined at an obtuse angle.

$∴\vec{GB}. \vec{GC} <0$

$⇒\left\{\vec b-\frac{\vec b+\vec c}{3}\right\}.\left\{\vec c-\frac{\vec b+\vec c}{3}\right\}<0$

$⇒(2\vec b-\vec c).(2\vec c-\vec b)<0$   ...(i)

$⇒5(\vec b.\vec c)-2|\vec b|^2-2|\vec c|^2<0$

We have,

$c=AB=|\vec b|,b=AC=|\vec c|$ and $a=BC=|\vec c-\vec b|$

$∴a^2=|\vec c-\vec b|^2=|\vec c|^2+|\vec b|^2-2(\vec b.\vec c)$

$⇒a^2=b^2-c^2-2(\vec b.\vec c)$

$⇒\vec b.\vec c=\frac{b^2+c^2-a^2}{2}$   ...(ii)

From (i) and (ii), we get

$5\frac{(b^2+c^2-a^2)}{2}<2c^2+2b^2 ⇒ b^2 + c^2 <5a^2$