if-x-y-gt-1-prove-x-2-y-1-y-2-x-1-8-

Question Number 78503 by jagoll last updated on 18/Jan/20

i f x, y > 1

p r o v e \frac{x^{2}}{y - 1} + \frac{y^{2}}{x - 1} ⩾ 8

Answered by ~blr237~ last updated on 18/Jan/20

We know that for all a, b \in R

{(a - b)}^{2} ⩾ 0 \Rightarrow 2 ab ⩽ a^{2} + b^{2} (*)

let take a = \frac{x}{\sqrt{y - 1}} b = \frac{y}{\sqrt{x - 1}}

a, b exist cause x > 1 and y > 1

Then (*) leads to \frac{2 xy}{\sqrt{(x - 1) (y - 1)}} ⩽ \frac{x^{2}}{y - 1} + \frac{y^{2}}{x - 1}

we also have

2 (2 x) ⩽ x^{2} + 4 \Rightarrow 4 ⩽ \frac{x^{2}}{x - 1} \Rightarrow 2 ⩽ \frac{x}{\sqrt{x - 1}}

samely we show 2 ⩽ \frac{y}{\sqrt{y - 1}}

Finaly we get

8 = 2 \times 2 \times 2 ⩽ 2 (\frac{x}{\sqrt{x - 1}}) (\frac{y}{\sqrt{y - 1}}) ⩽ \frac{x^{2}}{y - 1} + \frac{y^{2}}{x - 1}

Commented by jagoll last updated on 18/Jan/20

t h a n k s y o u s i r

Answered by behi83417@gmail.com last updated on 18/Jan/20

let : F (x, y) = \frac{x^{2}}{y - 1} + \frac{y^{2}}{x - 1}

\frac{\partial F}{\partial x} = \frac{2 x}{y - 1} - \frac{y^{2}}{{(x - 1)}^{2}}, \frac{\partial F}{\partial y} = \frac{- x^{2}}{{(y - 1)}^{2}} + \frac{2 y}{x - 1}

\Rightarrow {\begin{cases} 2 x {(x - 1)}^{2} - y^{2} (y - 1) = 0 \\ 2 y {(y - 1)}^{2} - x^{2} (x - 1) = 0 \end{cases} \Rightarrow

\Rightarrow {\begin{cases} {2 x}^{3} - {4 x}^{2} + 2 x - y^{3} + y^{2} = 0 (i) \\ {2 y}^{3} - {4 y}^{2} + 2 y - x^{3} + x^{2} = 0 (ii) \end{cases}

\Rightarrow {\begin{cases} x^{3} + y^{3} - 3 (x^{2} + y^{2}) + 2 (x + y) = 0 \\ 3 (x^{3} - y^{3}) - 5 (x^{2} - y^{2}) + 2 (x - y) = 0 \end{cases}

\overset{\frac{x}{y} = m}{\Rightarrow} {\begin{cases} m^{3} + 1 - 3 (m^{2} + 1) + 2 (m + 1) = 0 \\ 3 (m^{3} - 1) - 5 (m^{2} - 1) + 2 (m - 1) = 0 \end{cases}

\Rightarrow {\begin{cases} m^{3} - {3 m}^{2} + 2 m = 0 \\ {3 m}^{3} - {5 m}^{2} + 2 m = 0 \end{cases} \Rightarrow m = 0

\Rightarrow {\begin{cases} m^{2} - 3 m + 2 = 0 \Rightarrow m = 1, 2 \\ {3 m}^{2} - 5 m + 2 = 0 \Rightarrow m = 1, \frac{2}{3} \end{cases}

m = 0 \Rightarrow x = 0 [not ok : x > 1]

1) m = 1 \Rightarrow x = y \Rightarrow 2 x {(x - 1)}^{2} - x^{2} (x - 1) = 0

\Rightarrow (x - 1) (x^{2} - 2 x) = 0 \Rightarrow x = 2 = y \Rightarrow F (2, 2) = 8

2) m = 2 \Rightarrow x = 2 y \Rightarrow 4 y {(2 y - 1)}^{2} - y^{2} (y - 1) = 0

\Rightarrow y [{16 y}^{2} - 16 y + 4 - y^{2} + y] = 0

\Rightarrow {15 y}^{2} - 15 y + 4 = 0 \Rightarrow ▴ = 225 - 16 \times 15 < 0

3) m = \frac{2}{3} \Rightarrow x = \frac{2}{3} y \Rightarrow \frac{4}{3} y {(\frac{2}{3} y - 1)}^{2} - y^{2} (y - 1) = 0

\Rightarrow y [{16 y}^{2} - 48 y + 36 - {27 y}^{2} + 27 y] = 0

\Rightarrow {11 y}^{2} + 21 y - 36 = 0 \Rightarrow ▴ = 21^{2} + 44 \times 36

\Rightarrow [y = \frac{- 21 \pm 45}{22} = \frac{12}{11}, - 3 \Rightarrow x = \frac{8}{11}, - 2]

\Rightarrow F (\frac{8}{11}, \frac{12}{11}) = \frac{\frac{144}{121}}{\frac{8}{11} - 1} + \frac{\frac{64}{121}}{\frac{12}{11} - 1} =

= \frac{144}{- 33} + \frac{64}{11} = \frac{64 \times 33 - 11 \times 144}{33} = 16 > 8

[(x, y) = (- 2, - 3) not ok : x, y > 1]

\Rightarrow mimF = 8

Answered by MJS last updated on 18/Jan/20

let p, λ > 0; x = 1 + p \land y = 1 + λ p

\frac{x^{2}}{y - 1} + \frac{y^{2}}{x - 1} < 8

\frac{(λ^{3} + 1) p^{2} + 2 (λ^{2} + 1) p + λ + 1}{λ p} < 8

p^{2} + \frac{2 (λ^{2} - 4 λ + 1)}{λ^{3} + 1} p + \frac{1}{λ^{2} - λ + 1} < 0

let p = q - \frac{λ^{2} - 4 λ + 1}{λ^{3} + 1}

q^{2} + \frac{9 λ {(λ - 1)}^{2}}{{(λ + 1)}^{2} {(λ^{2} - λ + 1)}^{2}} < 0

but q^{2} > 0 \land λ > 0 \Rightarrow no solution

\Rightarrow

\frac{x^{2}}{y - 1} + \frac{y^{2}}{x - 1} < 8 is wrong

\Rightarrow

\frac{x^{2}}{y - 1} + \frac{y^{2}}{x - 1} ⩾ 8

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