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(a-b)^2 Expansion and Visualisation

If a 2 + b2...

Question

If $a^{2} + b^{2} + c^{2} = 1$ , $x^{2} + y^{2} + z^{2} = 1$ where a, b, c, x, y, z are real, prove that $a x + b y + c z \leq 1$

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Solution

Applying $A . M \geq G . M$ inequality,we have
$a^{2} + x^{2} \geq 2 a x$
$b^{2} + y^{2} \geq 2 b y$
$c^{2} + z^{2} \geq 2 c z$
Adding, $a^{2} + b^{2} + c^{2} + x^{2} + y^{2} + z^{2} \geq 2 (9 x + b y + c z)$ .Using the given conditions,it reduces to
$2 \geq 2 (a x + b y + c z) \Rightarrow a x + b y + c z \leq 1$

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a^{2} + b^{2} + c^{2} = 1

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a, b, c, x, y, z \geq 0,

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\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

\frac{- x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

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, then maximum value of ax + by + cz is

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