First slide

Introduction to real valued functions

Question

$If x, y and z are real and different and u = x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y, then u is always$

Moderate

A

non negative
B

zero
C

non positive
D

none of these

Solution

$\begin{array}{l} u = x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y \\ = \frac{1}{2} [2 x^{2} + 8 y^{2} + 18 z^{2} - 12 y z - 6 z x - 4 x y] \\ = \frac{1}{2} [(x^{2} - 4 x y + 4 y^{2}) + (4 y^{2} + 9 z^{2} - 12 y z) + (x^{2} + 9 z^{2} - 6 z x)] \\ = \frac{1}{2} [(x - 2 y)^{2} + (2 y - 3 z)^{2} + (3 z - x)^{2}] \geq 0 \end{array}$
Thus, u is always non-negative.

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