(1/x) + (1/y) + (1/z) = 1 find the minimum value of x^2 + y^2 + z^2


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Question Number 192612 by York12 last updated on 22/May/23

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ $f i n d t h e m i n i m u m v a l u e o f x^{2} + y^{2} + z^{2}$
Commented by Frix last updated on 23/May/23

$Yes of course!$
Commented by deleteduser1 last updated on 23/May/23

$F o r p o s i t i v e x, y, z; m i n (x^{2} + y^{2} + z^{2}) = 27 a t$ $x = y = z = 3$
Commented by York12 last updated on 23/May/23

$e x a c t l y$
	Answered by Subhi last updated on 23/May/23

	$A p p l y {t i t u}^{'} s L e m m a e n q u a l i t y$ $1 = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ⩾ \frac{9}{x + y + z} ⩾ \frac{9}{\sqrt{3 (x^{2} + y^{2} + z^{2})}}$ $\sqrt{3 (x^{x} + y^{2} + z^{2})} ⩾ 9$ $x^{2} + y^{2} + z^{2} ⩾ \frac{81}{3} = 27$ $t h e n, t h e m i n i v a l u e i s 27$
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