<s>[INST] 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
 [/INST] Triangle inequality  $$\Rightarrow\left | a-b \right |<c \Rightarrow \sum_{a,b,c}(a-b)^{2}<\sum_{a,b,c}c^{2}\Rightarrow 2\sum_{a,b,c}a^{2}-2\sum_{a,b,c}ab<\sum_{a,b,c}a^{2}\Rightarrow \frac{1}{2}<\frac{\sum\limits_{a,b,c}ab}{\sum\limits_{a,b,c}a^{2}}=x$$ $x\in \mathbb{R}\Rightarrow x^{2}\geq 0$ 
$$\Rightarrow(a-b)^2\geq 0\Rightarrow \sum_{a,b,c}(a-b)^2\geq 0\Rightarrow 2\sum_{a,b,c}a^2-2\sum_{a,b,c}ab\geq 0\Rightarrow x=\frac{\sum\limits_{a,b,c}ab}{\sum\limits_{a,b,c}a^{2}}\leq 1$$
Answer:
The true statements are:
(a) $1/2 ≤ x ≤  2$
(b) $1/2 ≤ x ≤  1$
(c) $1/2 < x ≤  1$
(c) $1/2 < x ≤  1$ gives the tightest bounds, but since the other two intervals are supersets of this interval, (a) and (b) are also satisfied. To further clarify:
$\forall\Delta ABC$, with sides a,b,c 
$x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}\Rightarrow 1/2 < x ≤  1\Rightarrow   
1/2 ≤ x ≤  1\Rightarrow 1/2 ≤ x ≤  2$
Also, $\forall x\in (\frac{1}{2},1],\exists  \Delta ABC$ with sides $ a,b,c \ni x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}$  since condition (c) gives the tightest bound.
However, $\exists x\in[\frac{1}{2},1],\forall\Delta ABC$ with sides $a,b,c ;\frac{ab + bc + ca}{a^2 + b^2 + c^2}\not=x$ and hence  $\exists x\in[\frac{1}{2},2],\forall\Delta ABC$ with sides $a,b,c ; \frac{ab + bc + ca}{a^2 + b^2 + c^2}\not=x.$ Thus the implication is one sided for conditions (a) and (b).
$\therefore x=\frac{ab + bc + ca}{a^2 + b^2 + c^2}\Leftrightarrow1/2 < x ≤  1 $but $1/2 ≤ x ≤  1\vee1/2 ≤ x ≤  2\not\Rightarrow \frac{ab + bc + ca}{a^2 + b^2 + c^2}=x $
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