if-x-y-8-x-y-R-prove-that-x-1-y-2-y-1-x-2-289-8-

Question Number 82375 by M±th+et£s last updated on 20/Feb/20

i f x + y = 8,, x, y \in R^{+}

p r o v e t h a t

{(x + \frac{1}{y})}^{2} + {(y + \frac{1}{x})}^{2} ⩾ \frac{289}{8}

Answered by MJS last updated on 21/Feb/20

f (x) = {(x + \frac{1}{8 - x})}^{2} + {(8 - x + \frac{1}{x})}^{2}

f (x) = \frac{2 (x^{2} - 8 x - 1) (x^{2} - 8 x + 32)}{x^{2} {(x - 8)}^{2}}

shift 4 to the left

x = t + 4 \Leftrightarrow t = x - 4

f (t) = \frac{2 {(t^{2} - 17)}^{2} (t^{2} + 16)}{{(t - 4)}^{2} {(t + 4)}^{2}}

f (t) has double zeros at t = \pm \sqrt{17} which also

are minima

and a local minimum at t = 0 with f (0) = \frac{289}{8}

\Rightarrow f (x) has double zeros at x = 4 \pm \sqrt{17} which

also are minima but 0 < x < 8 so they

are outside the given interval

and a local minimum at x = 4 with f (4) = \frac{289}{8}

{what}^{'} s left to show is that f (x) ⩾ \frac{289}{8} at the

borders : but \lim_{x \to 0} f (x) = \lim_{x \to 8} f (x) = + \infty \Rightarrow

\Rightarrow proven

Answered by mind is power last updated on 21/Feb/20

x^{2} + y^{2} + \frac{1}{y^{2}} + \frac{1}{x^{2}} + 2 \frac{x}{y} + 2 \frac{y}{x}

= (x^{2} + y^{2}) (1 + \frac{1}{x^{2} y^{2}} + \frac{2}{{xy}^{}})

, x^{2} + y^{2} ⩾ \frac{{(x + y)}^{2}}{2} = \frac{64}{2} = \frac{256}{8}

A M G M \Rightarrow

x y ⩽ \frac{{(x + y)}^{2}}{4} = 16

\Rightarrow \frac{1}{x y} ⩾ \frac{1}{16} \Rightarrow

(1 + \frac{2}{x y} + \frac{1}{x^{2} y^{2}}) ⩾ 1 + \frac{2}{16} + \frac{1}{256} = \frac{1 + 32 + 256}{256} = \frac{289}{256}

(x^{2} + y^{2}) (1 + \frac{2}{x y} + \frac{1}{x^{2} y^{2}}) ⩾ \frac{289}{256} . \frac{256}{8} = \frac{289}{8}

\Leftrightarrow {(x + \frac{1}{y})}^{2} + {(y + \frac{1}{x})}^{2} ⩾ \frac{289}{8}