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.
 A: 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:
$$a^2+b^2+c^2 > 2(a^2+b^2+c^2 - ab - ac - bc)\\ = 3(a^2+b^2+c^2) - (a+b+c)^2.$$
A: Using triangle inequality,
$a(b+c)+b(c+a)+c(a+b) \gt a^2 + b^2 + c^2$
$a^2+b^2+c^2+a(b+c)+b(c+a)+c(a+b) \gt 2 (a^2 + b^2 + c^2)$
$(a+b+c)^2 \gt 2(a^2 + b^2 + c^2)$
$\cfrac{a^2 + b^2 + c^2}{(a+b+c)^2} \lt \cfrac{1}{2}$
A: We need to prove that:$$2(a^2+b^2+c^2)<(a+b+c)^2,$$ which is true because
$$(a+b+c)^2-2(a^2+b^2+c^2)=\sum_{cyc}(2ab-a^2)=$$
$$=\sum_{cyc}(ab+ac-a^2)=\sum_{cyc}a(b+c-a)>0.$$
A: We want to show :
$$2(ab+bc+ca)>a^2+b^2+c^2$$
With :
$$a+b-c\geq0$$
$$a+c-b\geq 0$$
$$c+b-a\geq 0$$
As the inequalities are homogenous we need to show $0<x<1$ and $0<y<1$:
$$(x^2+y^2+1)<2(xy+x+y)$$
With :
$$x+y-1\geq0$$
$$x+1-y\geq 0$$
$$y+1-x\geq 0$$
Or :
$$(x^2+y^2+x+y)<2(xy+x+y)$$
Or :
$$2(x+y)<2(xy+x+y)$$
Done !
