<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] 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 !
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