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

Question

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.

Answer 1

Using triangle inequality,

$a(b+c)+b(c+a)+c(a+b) \gt a^2 + b^2 + c^2$

$a^2+b^2+c^2+a(b+c)+b(c+a)+c(a+b) \gt 2 (a^2 + b^2 + c^2)$

$(a+b+c)^2 \gt 2(a^2 + b^2 + c^2)$

$\cfrac{a^2 + b^2 + c^2}{(a+b+c)^2} \lt \cfrac{1}{2}$

Answer 2

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 !

Answer 3

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:

$$a^2+b^2+c^2 > 2(a^2+b^2+c^2 - ab - ac - bc)\\ = 3(a^2+b^2+c^2) - (a+b+c)^2.$$

Answer 4

We need to prove that:$$2(a^2+b^2+c^2)<(a+b+c)^2,$$ which is true because $$(a+b+c)^2-2(a^2+b^2+c^2)=\sum_{cyc}(2ab-a^2)=$$ $$=\sum_{cyc}(ab+ac-a^2)=\sum_{cyc}a(b+c-a)>0.$$