Hard inequality with x, y and z

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bel.jad5

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bel.jad5

#1 h Oct 21, 2017, 9:23 PM • 2 Y

Y by Adventure10, Mango247

Let $x$ , $y$ and $z$ positive real numbers such that $x+y+z+2=xyz$ . Find min and max of the following expression:
$\frac{x+1}{x^2+2}+\frac{y+1}{y^2+2}+\frac{z+1}{z^2+2}$

This post has been edited 5 times. Last edited by bel.jad5, Oct 22, 2017, 7:58 PM

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bel.jad5

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bel.jad5

#2 h Oct 22, 2017, 4:53 PM • 2 Y

Y by Adventure10, Mango247

No one has tried this one yet?

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leonardg

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leonardg

#4 h Oct 22, 2017, 8:07 PM • 1 Y

Y by Adventure10

Not only tried,but solved .Though,my strategy is unusual .
Congratulations .
1/2<E<=3/2 .

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ali3985

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ali3985

#5 h Oct 22, 2017, 8:19 PM • 2 Y

Y by Adventure10, Mango247

DELETED...

This post has been edited 2 times. Last edited by ali3985, Oct 22, 2017, 8:34 PM

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leonardg

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leonardg

#6 h Oct 22, 2017, 8:21 PM • 2 Y

Y by Adventure10, Mango247

Edited .

This post has been edited 1 time. Last edited by leonardg, Oct 22, 2017, 8:37 PM

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ali3985

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ali3985

#7 h Oct 22, 2017, 8:34 PM • 3 Y

Y by leonardg, Adventure10, Mango247

Sorry it was wrong you are right Mr Leonardg

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bel.jad5

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bel.jad5

#8 h Oct 22, 2017, 8:34 PM • 2 Y

Y by Adventure10, Mango247

leonardg wrote:

Not only tried,but solved .Though,my strategy is unusual .
Congratulations .
1/2<E<=3/2 .

Looking forward to see the "unusual" technique...it seems that the harder I try to make my inequalities, the more elegant and beautiful are the solutions

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bel.jad5

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bel.jad5

#9 h Oct 22, 2017, 8:35 PM • 2 Y

Y by Adventure10, Mango247

ali3985 wrote:

Sorry it was wrong you are right Mr Leonardg

Not a big deal...you are still a very good problem solver

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ali3985

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ali3985

#10 h Oct 22, 2017, 8:36 PM • 2 Y

Y by Adventure10, Mango247

Thank you Mr bel.jad5

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leonardg

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leonardg

#13 h Oct 22, 2017, 8:40 PM • 2 Y

Y by Adventure10, Mango247

bel.jad5 wrote:

Looking forward to see the "unusual" technique...it seems that the harder I try to make my inequalities, the more elegant and beautiful are the solutions

I started with this .

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bel.jad5

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bel.jad5

#14 h Oct 22, 2017, 8:52 PM • 2 Y

Y by Adventure10, Mango247

leonardg wrote:

bel.jad5 wrote:

Looking forward to see the "unusual" technique...it seems that the harder I try to make my inequalities, the more elegant and beautiful are the solutions

I started with this .

I used the same variable $x$ , $y$ and $z$ in my solution...

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leonardg

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leonardg

#15 h Oct 22, 2017, 8:55 PM • 1 Y

Y by Adventure10

Ok,but after my mentioned step,all is changed in the second .
And I get at least an isomorphical expression with the initial one .
Math is full of surprises .

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ali3985

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ali3985

#16 h Oct 22, 2017, 8:58 PM • 3 Y

Y by bel.jad5, Adventure10, Mango247

The condition is equivalent to

$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$
Then since $(x^2+2)(4+2) \ge 4(x+1)^2$

$\frac{x+1}{x^2+2}+\frac{y+1}{y^2+2}+\frac{z+1}{z^2+2} \le \frac{3}{2}(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1})=\frac{3}{2}$

And since $\frac{x+1}{x^2+2} > \frac{1}{2(x+1)} \iff x^2+4x > 0$

$\frac{x+1}{x^2+2}+\frac{y+1}{y^2+2}+\frac{z+1}{z^2+2} > \frac{1}{2}(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1})=\frac{1}{2}$

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bel.jad5

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bel.jad5

#17 h Oct 22, 2017, 9:00 PM • 1 Y

Y by Adventure10

ali3985 wrote:

The condition is equivalent to

$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$
Then since $(x^2+2)(4+2) \ge 4(x+1)^2$

$\frac{x+1}{x^2+2}+\frac{y+1}{y^2+2}+\frac{z+1}{z^2+2} \le \frac{3}{2}(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1})=\frac{3}{2}$

And since $\frac{x+1}{x^2+2} > \frac{1}{2(x+1)} \iff x^2+4x > 0$

$\frac{x+1}{x^2+2}+\frac{y+1}{y^2+2}+\frac{z+1}{z^2+2} > \frac{1}{2}(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1})=\frac{1}{2}$

Yes indeed...

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