Question

If a, b and c are distinct positive real numbers and \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1\], then ab + bc + ca is –
(a) less than 1
(b) equal to 1
(c) greater than 1
(d) any real number

Answer 1

Hint: First take an algebraic expression involving both the given expressions to solve the value of the required one. Use the condition that the square of any number is always greater than or equal to 0. By this get the condition on ab + bc + ca which is the required result.

Complete step-by-step answer:
The square of a number is always greater than or equal to zero. By using this condition, we can write the equation as,
\[{{\left( a+b+c \right)}^{2}}\ge 0\]
The given value in the question can be written in the form of
\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1.....\left( i \right)\]
By expanding the above inequality, we can write it as,
\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca\ge 0\]
We got this equation by basic algebraic identity given by:
\[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca\]
By substituting equation (i) in the inequality, we get it as,
\[1+2ab+2bc+2ca\ge 0\]
By subtracting 1 from both the sides, we get the inequality as,
\[2ab+2bc+2ca\ge -1\]
By dividing with 2 on both the sides, we get it as
\[ab+bc+ca\ge \dfrac{-1}{2}\]
Now, taking another inequality, we can write it as, \[{{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}}\ge 0\] because each of them is a square of a number.
By using the basic algebraic identity of the difference given by:
\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
By substituting this into our inequality, we can write it as
\[{{a}^{2}}-2ab+{{b}^{2}}+{{b}^{2}}-2bc+{{c}^{2}}+{{c}^{2}}-2ac+{{a}^{2}}\ge 0\]
By bringing similar terms together, we can write it as,
\[\left( {{a}^{2}}+{{a}^{2}} \right)+\left( {{b}^{2}}+{{b}^{2}} \right)+\left( {{c}^{2}}+{{c}^{2}} \right)-2ab-2bc-2ca\ge 0\]
By simplifying we can write it as,
\[2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca\ge 0\]
By dividing 2 on both the sides, we get,
\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca\ge 0\]
By subtracting equation (i), we get,
\[1-ab-bc-ca\ge 0\]
By simplifying, we can say the inequality is of the form
\[ab+bc+ca\le 1\]
Hence, option (a) is the right answer.

Note: First condition is just to prove that greater than 1 is not possible at all. Now we have the total range, so you can keep your answer confidently. Then the ideas are just to generalize the term ab + bc + ca. One is taken with a negative sign and other with a positive sign to get the maximum, minimum, or both.