x-y-z-R-2x-3y-4z-1-1-x-1-y-1-z-smallest-integer-value-

Question Number 115951 by Fikret last updated on 29/Sep/20

x, y, z ϵ R^{+}

2 x + 3 y + 4 z = 1 \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} s m a l l e s t i n t e g e r v a l u e ?

Answered by 1549442205PVT last updated on 30/Sep/20

From the hypothesis we have

P = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{2 x + 3 y + 4 z}{x} + \frac{2 x + 3 y + 4 z}{y}

+ \frac{2 x + 3 y + 4 z}{z} = 2 + 3 + 4 + 3 \frac{y}{x} + 2 \frac{x}{y}

+ 4 \frac{z}{x} + 2 \frac{x}{z} + 3 \frac{y}{z} + 4 \frac{z}{y}

⩾ 2 + 3 + 4 + 3 + 2 + 4 + 2 + 3 + 4 = 27

(Since x, yz > 0, \frac{x}{y}, \frac{y}{x}, \frac{y}{z}, \frac{z}{y}, \frac{x}{z}, \frac{z}{x} has

least integeral value equal to 1)

The equality ocurrs if and only if

{\begin{cases} 2 x + 3 y + 4 z = 1 \\ x = y = z \end{cases} \Leftrightarrow x = y = z = \frac{1}{9}

Thus, the least integral value of P is

{(\frac{1}{x} + \frac{1}{y} + \frac{1}{z})}_{\min} = 27 when

x = y = z = 1 / 9

Commented by soumyasaha last updated on 01/Oct/20

Assuming x, y, z positive, we have

\frac{2 x + 3 y + 4 z}{3} ⩾ \sqrt[3]{2 x . 3 y . 4 z}

\Rightarrow 24 xyz ⩽ \frac{1}{27}

\Rightarrow \frac{1}{xyz} ⩾ 27.24 \dots \dots \dots . (i)

Again, \frac{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}{3} ⩾ \sqrt[3]{\frac{1}{x} . \frac{1}{y} . \frac{1}{y}}

\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ⩾ 3 . \sqrt[3]{\frac{1}{xyz}}

\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ⩾ 3 . \sqrt[3]{27.24}

\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ⩾ 18 . \sqrt[3]{3}

∴ least integral value is 26