let x,y,z be positive real numbers such that x+y+z=1. Determine the minimum value of (1/x)+(4/y)+(9/z).


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Question Number 119303 by bobhans last updated on 23/Oct/20

$l e t x, y, z b e p o s i t i v e r e a l n u m b e r s$ $s u c h t h a t x + y + z = 1 . D e t e r m i n e$ $t h e m i n i m u m v a l u e o f \frac{1}{x} + \frac{4}{y} + \frac{9}{z} .$
	Answered by TANMAY PANACEA last updated on 23/Oct/20

	$\frac{1}{x} + \frac{4}{y} + \frac{9}{z}$ $\frac{1^{2}}{x} + \frac{2^{2}}{y} + \frac{3^{2}}{z} ⩾ \frac{{(1 + 2 + 3)}^{2}}{x + y + z}$ $s o m i n i m u m v a l u e = 36$ $u s i n g T i t t u l e m m a$
	Answered by bobhans last updated on 24/Oct/20

	$b y C h a u c h y - S c h w a r z i n e q u a l i t y$ $\frac{1}{x} + \frac{4}{y} + \frac{9}{z} = (x + y + z) (\frac{1}{x} + \frac{4}{y} + \frac{9}{z}) ⩾ {(1 + 2 + 3)}^{2} = 36$ $e q u a l i t y h o l d i f o n l y i f (x, y, z) = (\frac{1}{6}, \frac{1}{3}, \frac{1}{2})$
	Answered by 1549442205PVT last updated on 24/Oct/20
	$Applying the inequality (ax+by+cz)^2 ≤(a^2 +b^2 +c^2 (x^2 +y^2 +z^2 ) we have (1+2+3)^2 =((1/( (√x))).(√x)+(2/( (√y))).(√y)+(3/z).(√z))^2 ≤((1/x)+(4/y)+(9/z))(x+y+z)=(1/x)+(2/y)+(3/z) ⇒(1/x)+(2/y)+(3/z)≥36.The equality ocurrs if and only if { (((1/x)=(2/y)=(3/z)=(6/(x+y+z))=6)),((x+y+z=1)) :} ⇔x=(1/6),y=(1/3),z=(1/2).Thus, ((1/x)+(2/y)+(3/z))_(min) =36 when x=(1/6),y=(1/3),z=(1/2)$
	$Applying the inequality$ ${(ax + by + cz)}^{2} ⩽ (a^{2} + b^{2} + c^{2} (x^{2} + y^{2} + z^{2})$ $we have$ ${(1 + 2 + 3)}^{2} = {(\frac{1}{\sqrt{x}} . \sqrt{x} + \frac{2}{\sqrt{y}} . \sqrt{y} + \frac{3}{z} . \sqrt{z})}^{2}$ $⩽ (\frac{1}{x} + \frac{4}{y} + \frac{9}{z}) (x + y + z) = \frac{1}{x} + \frac{2}{y} + \frac{3}{z}$ $\Rightarrow \frac{1}{x} + \frac{2}{y} + \frac{3}{z} ⩾ 36 . The equality ocurrs$ $if and only if {\begin{cases} \frac{1}{x} = \frac{2}{y} = \frac{3}{z} = \frac{6}{x + y + z} = 6 \\ x + y + z = 1 \end{cases}$ $\Leftrightarrow x = \frac{1}{6}, y = \frac{1}{3}, z = \frac{1}{2} . Thus,$ ${(\frac{1}{x} + \frac{2}{y} + \frac{3}{z})}_{\min} = 36 when$ $x = \frac{1}{6}, y = \frac{1}{3}, z = \frac{1}{2}$
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