Question

If $ {\alpha ^2} + {\beta ^2} + {\gamma ^2} = 1 $ , then highest integral value of $ \alpha \beta + \beta \gamma + \gamma \alpha $ is
A.0
B.1
C.2
D.3

Answer 1

Hint: In this question, we need to find highest integral value of $ \alpha \beta + \beta \gamma + \gamma \alpha $ , such that $ {\alpha ^2} + {\beta ^2} + {\gamma ^2} = 1 $ . For this, we will use identity $ {\left( {\alpha + \beta + \gamma } \right)^2} = {\alpha ^2} + {\beta ^2} + {\gamma ^2} + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) $ and relation the concept that the arithmetic mean is always greater than or is equals to the geometric mean. Then, find the range in which $ \alpha \beta + \beta \gamma + \gamma \alpha $ lies and the highest integral value of that range will be our required answer.

Complete step-by-step answer:
We know that $ {\left( {\alpha + \beta + \gamma } \right)^2} = {\alpha ^2} + {\beta ^2} + {\gamma ^2} + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) - - - - (i) $
It is given, $ {\alpha ^2} + {\beta ^2} + {\gamma ^2} = 1 - - - - (ii) $
Therefore, substituting the value of equation (ii) in the equation (i) we get
\[
\Rightarrow {\left( {\alpha + \beta + \gamma } \right)^2} = {\alpha ^2} + {\beta ^2} + {\gamma ^2} + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) \\
   = 1 + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) - - - - (iii) \\
 \]
We know that $ \alpha + \beta + \gamma \geqslant 0 $ for all real $ \alpha ,\beta ,\gamma $
Therefore, equation (iii) can be written as:
\[
\Rightarrow {\left( {\alpha + \beta + \gamma } \right)^2} \geqslant 0 \\
  1 + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) \geqslant 0 - - - - (iv) \\
 \]
Equation (iv) can be re-written as:
\[
\Rightarrow 1 + 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) \geqslant 0 \\
  \alpha \beta + \beta \gamma + \gamma \alpha \geqslant \dfrac{{ - 1}}{2} - - - - (v) \\
 \]
We know that the Arithmetic Mean of any two real numbers is always greater than their Geometric Mean i.e $ \dfrac{{a + b}}{2} \geqslant \sqrt {ab} - - - - (vi) $ .
Let $ a = {\alpha ^2} $ and $ b = {\beta ^2} $ .
Substitute the values of ‘a’ and ‘b’ in the equation (vi), we get
 $
\Rightarrow \dfrac{{a + b}}{2} \geqslant \sqrt {ab} \\
\Rightarrow \dfrac{{{\alpha ^2} + {\beta ^2}}}{2} \geqslant \sqrt {{\alpha ^2} \times {\beta ^2}} \\
\Rightarrow {\alpha ^2} + {\beta ^2} \geqslant 2\alpha \beta - - - - (vii) \\
  $
Similarly,
 $ {\alpha ^2} + {\gamma ^2} \geqslant 2\alpha \gamma - - - - (viii) $
And,
\[{\beta ^2} + {\gamma ^2} \geqslant 2\beta \gamma - - - - (ix)\]
Adding equations (vii), (viii) and (ix), we get
\[
\Rightarrow 2\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2}} \right) \geqslant 2\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) \\
\Rightarrow \left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) \leqslant 1 - - - - (x) \\
 \]
From equations (v) and (x), we have
 $ \Rightarrow \dfrac{{ - 1}}{2} \leqslant \left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) \leqslant 1 $
Therefore, \[\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right)\] lies in interval $ \left[ {\dfrac{{ - 1}}{2},1} \right] $ .
Hence, the highest integral value is 1.
So, the correct answer is “Option B”.

Note: It is interesting to note that the value of the arithmetic mean is always greater or equals to the geometric mean. Arithmetic mean is the average value of the quantities involved in the calculations while the geometric mean is the square root of the product of the quantities involved in the calculations.