<s>[INST] 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.
 [/INST] Using triangle inequality,
$a(b+c)+b(c+a)+c(a+b) \gt a^2 + b^2 + c^2$
$a^2+b^2+c^2+a(b+c)+b(c+a)+c(a+b) \gt 2 (a^2 + b^2 + c^2)$
$(a+b+c)^2 \gt 2(a^2 + b^2 + c^2)$
$\cfrac{a^2 + b^2 + c^2}{(a+b+c)^2} \lt \cfrac{1}{2}$
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