Which Values of x and y Satisfy the Inequality x^2y+y^2x > 6?

[evagelos]

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?

 

[HallsofIvy]

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 [tex]x^2y+ y^2x= 6[/itex]. You can think of that as a quadratic equation in y and solve using the quadratic formula:
$$y= \frac{-x^2\pm\sqrt{x^4+ 24x}}{2x}$$

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".

 

[evagelos]

The best way to handle complicated inequalities is to look first at the associated equality. 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:
$$y= \frac{-x^2\pm\sqrt{x^4+ 24x}}{2x}$$

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

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

 

[CellCoree]

To determine the values of x and y that satisfy this inequality, we need to find the critical points where the inequality changes from being greater than to being less than 6. To do this, we can set the left side of the inequality equal to 6 and solve for x and y.

xy(x+y) = 6

We can then use techniques such as substitution or factoring to solve for the values of x and y that satisfy this equation. For example, if we substitute x = 1 and y = 2, we get:

1(2)(1+2) = 6

2(3) = 6

6 = 6

Therefore, the values of x = 1 and y = 2 satisfy this inequality. Similarly, we can try other values of x and y to see if they also satisfy the inequality. For instance, when x = 2 and y = 1, we get:

2(1)(2+1) = 6

2(3) = 6

6 = 6

Therefore, the values of x = 2 and y = 1 also satisfy the inequality.

In conclusion, the inequality x^2y+y^2x > 6 is satisfied when x = 1 and y = 2, or when x = 2 and y = 1. These are the critical points where the inequality changes from being greater than 6 to being less than 6. Any other values of x and y will either make the left side of the inequality greater than 6 or less than 6.