If $a^{2} + b^{2} + c^{2} = 1$ and $x^{2} + y^{2} + z^{2} = 1$ , show that $a x + b y + c z < 1$ .

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Solution

This is true only if $a, b, c, x, y, z$ are positive numbers.
Given $a^{2} + b^{2} {+ c}^{2} = 1, x^{2} + y^{2} {+ z}^{2} = 1$
Apply $A M \geq G M$ for $a^{2}, x^{2}$
$⟹ \frac{a^{2} + x^{2}}{2} \geq a x \to$ eqn 1
Apply $A M \geq G M$ for $b^{2}, y^{2}$
$⟹ \frac{b^{2} + y^{2}}{2} \geq b y \to$ eqn 2
Apply $A M \geq G M$ for $c^{2}, z^{2}$
$⟹ \frac{c^{2} + z^{2}}{2} \geq c z \to$ eqn 3
Adding corresponding sides of eqns 1,2,3
$\frac{a^{2} + x^{2}}{2} + \frac{b^{2} + y^{2}}{2} + \frac{c^{2} + z^{2}}{2} \geq a x + b y + c z ⟹ \frac{a^{2} + x^{2} + b^{2} + y^{2} + c^{2} + z^{2}}{2} \geq a x + b y + c z ⟹ \frac{1 + 1}{2} \geq a x + b y + c z ⟹ a x + b y + c z \leq 1$

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