In triangle ABC, range of a2+b2+c2ab+bc+ca is (a,b,c are sides of triangle)

a2+b2+c2ab+bc+ac(a-b)2+(b-c)2+(c-a)2≥0 a2+b2+c2≥ab+bc+ca1≤a2+b2+c2ab+bc+caa+b>c⇒c-b a⇒a-c b⇒b-a<c(c-b)2+(a-c)2+(b-a)2<a2+b2+c22(a2+b2+c2)-2(ab+bc+ca)<a2+b2+c2a2+b2+c2ab+bc+ca<2 Range of a2+b2+c2ab+bc+cais [1,2)

In triangle ABC, range of a2+b2+c2ab+bc+ca is (a,b,c are sides of triangle)

a2+b2+c2ab+bc+ac(a-b)2+(b-c)2+(c-a)2≥0 a2+b2+c2≥ab+bc+ca1≤a2+b2+c2ab+bc+caa+b>c⇒c-b a⇒a-c b⇒b-a<c(c-b)2+(a-c)2+(b-a)2<a2+b2+c22(a2+b2+c2)-2(ab+bc+ca)<a2+b2+c2a2+b2+c2ab+bc+ca<2 Range of a2+b2+c2ab+bc+cais [1,2)

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In triangle ABC, range of $\frac{a^{2} + b^{2} + c^{2}}{ab + bc + ca}$ is (a,b,c are sides of triangle)

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(1,2]

(1,2)

[1,2)

[1,2]

answer is B.

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Detailed Solution

$\begin{array}{rcl} \frac{a^{2} + b^{2} + c^{2}}{ab + bc + ac} \\ {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} & \geq & 0 a^{2} + b^{2} + c^{2} \geq ab + bc + ca \\ 1 & \leq & \frac{a^{2} + b^{2} + c^{2}}{ab + bc + ca} \\ a + b & > & c \Rightarrow c - b < a \\ b + c & > & a \Rightarrow a - c < b \\ c + a & > & b \Rightarrow b - a < c \\ {(c - b)}^{2} + {(a - c)}^{2} + {(b - a)}^{2} & < & a^{2} + b^{2} + c^{2} \\ 2 (a^{2} + b^{2} + c^{2}) - 2 (ab + bc + ca) & < & a^{2} + b^{2} + c^{2} \\ \frac{a^{2} + b^{2} + c^{2}}{ab + bc + ca} & < & 2 \end{array}$
$\begin{array}{rcl} Range of \frac{a^{2} + b^{2} + c^{2}}{ab + bc + ca} is [1, 2) \end{array}$