First slide

Introduction to real valued functions

Question

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

Moderate

A

$[\frac{1}{2}, 2]$
B

$[- 1, 2]$
C

$[- \frac{1}{2}, 1]$
D

$[- 1, \frac{1}{2}]$

Solution

$Given that a^{2} + b^{2} + c^{2} = 1 - - - - (1)$
$\begin{array}{l} we know that (a + b + c)^{2} \geq 0 \\ \Rightarrow a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a \geq 0 \end{array}$
$\Rightarrow 2 (a b + b c + c a) \geq - 1 [Using (1)]$
$\Rightarrow a b + b c + c a \geq - 1 / 2 - - - - - (2)$
$Also we know that, \frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] \geq 0$
$\Rightarrow a^{2} + b^{2} + c^{2} - a b - b c - c a \geq 0$
$\Rightarrow a b + b c + c a \leq 1 [u \sin g (1)]$
$\Rightarrow a b + b c + c a \leq 1 - - - - (3)$
$\begin{array}{l} Combining (2) and (3), we get \\ - 1 / 2 \leq a b + b c + c a \leq 1 \\ \Rightarrow a b + b c + c a \in [- \frac{1}{2}, 1] \end{array}$

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