# X^2y+y^2x >6 inequality

## Main Question or Discussion Point

For what values of x and , y is the following inequality satisfied:

$$x^2y+y^2x >6$$

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

xy(x+y)>6 then what?

## Answers and Replies

HallsofIvy
Homework Helper

For what values of x and , y is the following inequality satisfied:

$$x^2y+y^2x >6$$

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

xy(x+y)>6 then what?
The best way to handle complicated inequalities is to look first at the associated equality. Here, we start by looking at $$x^2y+ y^2x= 6[/itex]. You can think of that as a quadratic equation in y and solve using the quadratic formula: [tex]y= \frac{-x^2\pm\sqrt{x^4+ 24x}}{2x}$$

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".

The best way to handle complicated inequalities is to look first at the associated equality. Here, we start by looking at $$x^2y+ y^2x= 6[/itex]. You can think of that as a quadratic equation in y and solve using the quadratic formula: [tex]y= \frac{-x^2\pm\sqrt{x^4+ 24x}}{2x}$$

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$$x^3y+yx^3>10$$ can we use graphic procedures??

uart

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 :

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

and

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

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