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-

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} = {(\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}