Let $a, b$ and $c$ be the lengths of the sides of an arbitrary triangle. Pick out the true statements. Let $a, b$ and $c$ be the lengths of the sides of an arbitrary triangle. Define
$$x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}.$$
Pick out the true statements.
(a) $1/2 ≤ x ≤  2$.
(b) $1/2 ≤ x ≤  1$.
(c) $1/2 < x ≤  1$.    
How can I able to solve this problem

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*We first rule out (a) by writing $(a-b)^2\geq 0$, $(b-c)^2\geq 0$, $(c-a)^2\geq 0 $ and adding side by side to get $2a^2+2b^2+2c^2\geq 2ab + 2bc+2ac$. This tells us $\frac{ab+bc+ac}{a^2+b^2+c^2}\leq 1 $. Therefore $x$ cannot be greater than 1. Therefore (a) is not your answer. 

*We now rule out (b) by writing: from the Cosine Law $a^2+b^2-2ab\cos\gamma=c^2$,  $b^2+c^2-2bc\cos\alpha=a^2$ and $c^2+a^2-2ca\cos\beta=b^2.$ Adding side by side gives $a^2+b^2+c^2=2ab\cos\gamma+2bc\cos\alpha+2ac\cos\beta<2ab+2bc+2ac.$ This implies $x=\frac{ab+bc+ac}{a^2+b^2+c^2}>\frac{1}{2}$. Notice the last inequality follows since the angles in a triangle add up to $180^o$.

