Write it as $\frac{(a+(b+c))^2}{bc}$ Now use $A.M$$\ge$$G.M$ So $b+c\ge2\sqrt{bc}$ Now our expression is $\frac{(a^2+(b+c)^2+2a(b+c)}{bc}$ Which is equivalent to 4+ $\frac{(a^2+2\sqrt{2}a)}{bc}$ Now since $b+c$>$a$ So $\frac{(a^2+2\sqrt{2}a)}{bc}$<$\frac{((b+c)^2+2\sqrt{2}(b+c))}{bc}$ Which is $4+8+4=16-1=15$ Hence,$T=15$