Show that $\frac 1 2 <\frac{ab+bc+ca}{a^2+b^2+c^2} \le 1$

Question

If $a,b,c$ are sides of a triangle, then show that $$\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2} \le 1$$

Trial: $$(a-b)^2+(b-c)^2+(c-a)^2 \ge 0\\\implies a^2+b^2+c^2 \ge ab+bc+ca $$ But how I prove $\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2}$ . Please help.

Answer 1

Adding the triangle inequalities:$$a^{2}>(b-c)^{2}\\b^{2}>(a-c)^{2}\\c^{2}>(b-a)^{2}$$

We get $$a^{2}+b^{2}+c^{2}>2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca$$

Now divide by $a^{2}+b^{2}+c^{2}$ to get,$$1>2-2\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)\\\rightarrow \frac{ab+bc+ca}{a^2+b^2+c^2}>\frac{1}{2}$$

Answer 2

We know from triangle inequality

In a triangle the sum of any two sides is greater than the third side, from which it follows that

$(a+b-c)(a+c-b)+(b+c-a)(b+a-c)+(c+a-b)(c+b-a)>0$

Just expand this to get the inequality

$\Rightarrow a^2-b^2-c^2+2bc+b^2-c^2-a^2+2ac+c^2-a^2-b^2+2ab>0$

$\Rightarrow -a^2-b^2-c^2+2(ab+bc+ca)>0$