Question

If $ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1, $ then $ ab+bc+ca $ lies in the interval

• A. $ \left[ \frac{1}{2},2 \right] $
• B. $ [-1,2] $
• C. $ \left[ -\frac{1}{2},1 \right] $
• D. $ \left[ -1,\frac{1}{2} \right] $

Answer 1

Answer: (C)

Given, $ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1 $ ...(i)
$ {{(a+b+c)}^{2}}\ge 0 $
$ \Rightarrow $ $ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(ab+bc+ca)\ge 0 $
$ \Rightarrow $ $ 1+2(ab+bc+ca)\ge 0 $
$ \Rightarrow $ $ ab+bc+ca\ge -\frac{1}{2} $ ....(ii) and
$ {{(a-b)}^{2}}\ge 0 $
$ \Rightarrow $ $ {{a}^{2}}+{{b}^{2}}-2ab\ge 0 $
$ \Rightarrow $ $ {{a}^{2}}+{{b}^{2}}\ge 2ab $ Similarly, $ {{b}^{2}}+{{c}^{2}}\ge 2bc,{{c}^{2}}+{{a}^{2}}\ge 2ca $
$ \therefore $ $ 2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\ge 2ab+2bc+2ca $
$ \Rightarrow $ $ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}\ge ab+bc+ca $
$ \Rightarrow $ $ 1\ge ab+bc+ca $ ...(iii)
Hence, from Eqs. (ii) and (iii),
$ -\frac{1}{2}\le ab+bc+ca\le 1 $
$ \therefore $ Interval $ =\left[ -\frac{1}{2},1 \right] $