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
 A: *

*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 $180^o$.

A: We know in every triangle $ABC$ there are some useful relations called Law of cosines:
$$a^2=b^2+c^2-2bc\cos(A)\\b^2=a^2+c^2-2ac\cos(B)\\c^2=a^2+b^2-2ab\cos(C)$$ By adding them we have: $$a^2+b^2+c^2=2(bc\cos(A)+ac\cos(B)+ab\cos(C))$$ and if we take $A=B=C=60^{~\text{o}}$ then $x=1$($ABC$ is a Equilateral).
Now take $A=90^{~\text{o}},B=45^{~\text{o}},C=45^{~\text{o}}$ and using  $b=a\sin(B), c=a\cos(B)$ we have: $$x=\frac{a^2\sin(B)+a^2\sin(B)\cos(B)+a^2\cos(B)}{a^2}\sim 0.9$$ It seems that $x\leq1$.
A: 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 $
A: because the expression is homogeneous of degree $0$ we may assume $a+b+c=1$ 
using $\sum$ to indicate symmetric sums we have, (with $\sum a=1)$:
$$\frac{\sum 2ab}{\sum a^2} = \frac {(\sum a)^2-\sum a^2}{\sum a^2} 
$$
thus
$$\frac{ab + bc + ca}{a^2 + b^2 + c^2} = \frac12 \left(\frac1{a^2 + b^2 + c^2}-1\right)
$$
this expression, if $a,b,c$ are the sides of a triangle, has a maximum value of $1$ when the triangle is equilateral, and an unattained minimum of $\frac12$ in the degenerate case where one of the sides is zero
added: apologies, as I see Rhaldryn has already posted a more thorough answer using a similar line of argument. I will leave my answer here because the expression is slightly different, and may be a useful complement.
