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160 posts

Bet667

#1 h Apr 19, 2025, 4:14 PM

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Let $x,y\ge 0$ such that $2(x+y)=1+xy$ then find minimal value of $x+\frac{1}{x}+\frac{1}{y}+y$

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160 posts

Bet667

#2 h Apr 19, 2025, 4:15 PM

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i think answer is 8 but i think my solution is wrong can you guys help me

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160 posts

Bet667

#3 h Apr 19, 2025, 5:24 PM

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anyone??

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#4 h Apr 19, 2025, 5:40 PM

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The answer should be 8. I used the condition to get $y=2+\frac{3}{x-2}$ and bashed with derivative, but there's probably a nicer solution.

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160 posts

Bet667

#5 h Apr 19, 2025, 5:43 PM

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Can u show me your sol pls

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#6 h Apr 19, 2025, 5:59 PM • 1 Y

Y by teomihai

Bet667 wrote:

Can u show me your sol pls

you can use the substitution $x+y=2u$ and $xy=v^2$ with $u \geq v\geq 0$ . first you get: $1+v^2 \geq 4v$ which gives the boundaries of $v$ . then you study the expression to maximise as a function of $v$ . which is straightforward. The minimum is $8$ .

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sqing

#7 h Apr 20, 2025, 2:35 AM

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Let $x,y\ge 0$ such that $2(x+y)=1+xy.$ Prove that $x+\frac{1}{x}+\frac{1}{y}+y\geq 8$ Equality holds when $x=y=2-\sqrt 3$ or $x=y=2+\sqrt 3 .$

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#8 h Apr 20, 2025, 4:39 AM • 1 Y

Y by aidan0626

Bet667 wrote:

Let $x,y\ge 0$ such that $2(x+y)=1+xy$ then find minimal value of $x+\frac{1}{x}+\frac{1}{y}+y$

because:
$x+\frac{1}{x}+\frac{1}{y}+y=x+y+\frac{1}{x}+\frac{1}{y} \geq \frac{4xy}{x+y}+\frac{4}{x+y} =\frac{4(xy+1)}{x+y} = 8$ Equality at $x=y$ and we get: $x=y=2+\sqrt 3$ or $x=y=2-\sqrt 3$ .

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sqing

#9 h Apr 20, 2025, 9:26 AM

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Let $x,y\geq 0$ such that $4(x+y)=12+xy.$ Prove that $x+y+\frac{1}{x}+\frac{1}{y}\geq5$ Let $x,y\geq 0$ such that $7(x+y)=24+xy.$ Prove that $x+y+\frac{1}{x}+\frac{1}{y}\geq5$ Let $x,y\geq 0$ such that $4(x+y)=15+xy.$ Prove that $x+y+\frac{1}{x}+\frac{1}{y}\geq \frac{20}{3}$ Let $x,y\geq 0$ such that $6(x+y)=27+xy.$ Prove that $x+y+\frac{1}{x}+\frac{1}{y}\geq \frac{20}{3}$

This post has been edited 1 time. Last edited by sqing, Apr 20, 2025, 9:39 AM

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#11 h Apr 20, 2025, 9:56 AM

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Let $x,y\geq 0$ such that $k(x+y)=l+xy.$ Prove that $x+y+\frac{1}{x}+\frac{1}{y} \geq 2(k-\sqrt{k^2-l}+\frac{k+\sqrt{k^2-l}}{l})$ Where $k^2>l.$ and $k,l \in N^+.$

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sqing

#12 h Apr 20, 2025, 10:01 AM

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Good.

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