- #1

evagelos

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[tex]x^2y+y^2x >6[/tex]

I tried to give a proof and i went as far:

xy(x+y)>6 then what?

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- Thread starter evagelos
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- #1

evagelos

- 315

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[tex]x^2y+y^2x >6[/tex]

I tried to give a proof and i went as far:

xy(x+y)>6 then what?

- #2

HallsofIvy

Science Advisor

Homework Helper

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The best way to handle complicated inequalities is to look first at the associated

[tex]x^2y+y^2x >6[/tex]

I tried to give a proof and i went as far:

xy(x+y)>6 then what?

[tex]y= \frac{-x^2\pm\sqrt{x^4+ 24x}}{2x}[/tex]

Graph that on, say, a graphing calculator and it shows the boundary between "> 6" and "< 6". Putting in one (x, y) point for each region will tell you which regions are "> 6".

- #3

evagelos

- 315

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The best way to handle complicated inequalities is to look first at the associatedequality. Here, we start by looking at [tex]x^2y+ y^2x= 6[/itex]. You can think of that as a quadratic equation in y and solve using the quadratic formula:

[tex]y= \frac{-x^2\pm\sqrt{x^4+ 24x}}{2x}[/tex]

Graph that on, say, a graphing calculator and it shows the boundary between "> 6" and "< 6". Putting in one (x, y) point for each region will tell you which regions are "> 6".

You mean then, that there is no solid proof for the inequality but only graphic procedures?

How about the inequality[tex]x^3y+yx^3>10[/tex] can we use graphic procedures??

- #4

uart

Science Advisor

- 2,797

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You mean then, that there is no solid proof for the inequality but only graphic procedures?

No that's not what Halls said. He found the boundary which precisely identifies the region of the x,y plane you are looking for. Graphing it is optional (though very helpful in my opinion).

For x>0 it reduces to the union of :

[tex]

y > \frac{-x^2 + \sqrt{x^4+ 24x}}{2x}

[/tex]

and

[tex]

y < \frac{-x^2 - \sqrt{x^4+ 24x}}{2x}

[/tex]

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