If a + b + c = 9 (where a, b, c are real numbers), then the minimum value of a^2 + b^2 + c^2 is (a) 81 (b) 100 (c) 9 (d) 27

Solution

To find the minimum value of \(a^{2}+b^{2}+c^{2}\) given that \(a+b+c=9\), where \(a, b, c\) are real numbers, we can follow these steps:

Step 1: Express \(a^{2}+b^{2}+c^{2}\) in Terms of \(a+b+c\)

We start by expanding \((a+b+c)^2\), which gives us:

\[
\begin{aligned}
(a+b+c)^2 & = a^2 + b^2 + c^2 + 2(ab + bc + ca) \\
\end{aligned}
\]

Thus, we can express \(a^{2}+b^{2}+c^{2}\) as:

\[
\begin{aligned}
a^2+b^2+c^2 & =(a+b+c)^2-2(ab+bc+ca) \\
& =9^2-2(ab+bc+ca)
\end{aligned}
\]

Step 2: Maximizing \(ab + bc + ca\)

Since \(a^2+b^2+c^2\) will be minimum if \(ab + bc + ca\) is maximum, we consider the condition for maximizing \(ab + bc + ca\).

This condition is met when \(a = b = c\), due to the symmetry of the expression and given that their sum is fixed (\(a + b + c = 9\)). Thus, when \(a = b = c = 3\), \(ab + bc + ca\) is maximized.

Step 3: Calculating the Minimum Value

Substituting \(a = b = c = 3\) into our expression:

\[
\begin{aligned}
a^2+b^2+c^2 & = 81 – 2(3 \times 3 + 3 \times 3 + 3 \times 3) \\
& = 81 – 2(27) \\
& = 81 – 54 \\
& = 27
\end{aligned}
\]

Therefore, the minimum value of \(a^2+b^2+c^2\) is 27.

The correct answer is (d) 27.