Given $a,b,c$ are sides of a triangle, Prove that :- $\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$ 
Given $a,b,c$ are sides of a triangle, Prove that :-
$$\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$$

What I Tried:- I was able to solve the left hand side inequality. From RMS-AM Inequality on $a,b,c$ :-
$$\sqrt{\frac{a^2+b^2+c^2}{3}} \geq \frac{a+b+c}{3}$$
$$\rightarrow \frac{a^2+b^2+c^2}{3} \geq \frac{(a+b+c)^2}{9}$$
$$\rightarrow \frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2}$$
I got no progress for the second part. I also have a clue, that $a,b,c$ are sides of a triangle, which I have not used yet. So maybe that should be used somehow, but I am not getting it.
Can anyone help me? Thank You.

Note that the sum of any two sides is greater then the third side, which shows $$0 < (a+b-c)(a-b+c) = a^2 - (b-c)^2.$$
Similarly $$\begin{eqnarray}
a^2 &>& (b-c)^2\\
b^2 &>& (a-c)^2\\
c^2 &>& (a-b)^2
\end{eqnarray}.$$  Summing these lines:
$$a^2+b^2+c^2 > 2(a^2+b^2+c^2 - ab - ac - bc)\\ = 3(a^2+b^2+c^2) - (a+b+c)^2.$$
