If $a,\ b,\ c$ are sides of triangle, then $\frac{(a + b + c)^{2}}{(ab + bc + ca)}$ always belongs to

Question

Dr. Abubacker · Accepted Answer

(c)
Since, $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} \geq 0$
$\Rightarrow \ \ 2\left( a^{2} + b^{2} + c^{2} \right) \geq 2(ab + bc + ca)$
$\Rightarrow \ \ \frac{a^{2} + b^{2} + c^{2}}{(ab + bc + ca)} \geq 1$
$\Rightarrow \ \frac{a^{2} + b^{2} + c^{2}}{ab + bc + ca} + 2 \geq 3$
Hence, option (c) is correct

Q. If $a,\ b,\ c$ are sides of triangle, then $\frac{(a + b + c)^{2}}{(ab + bc + ca)}$ always belongs to

Explanation

Reviewed by Dr. Abubacker

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