# find minimum possible value..

• Aug 11th 2009, 09:50 PM
nh149
find minimum possible value..
can anyone help me with this problem?
find the minimum possible value of x^2 + y^2 given that x,y are real numbers such that

xy(x^2 - y^2 ) = x^2 + y^2 , x is not equal to 0.

thanx(Talking)
• Aug 12th 2009, 01:02 AM
Hello nh149
Quote:

Originally Posted by nh149
can anyone help me with this problem?
find the minimum possible value of x^2 + y^2 given that x,y are real numbers such that

xy(x^2 - y^2 ) = x^2 + y^2 , x is not equal to 0.

thanx(Talking)

I don't know whether this works, but have you tried the substitution $\displaystyle y = xz$, since the expressions are homogeneous?

Then $\displaystyle x^2+y^2 = x^2(1+z^2)$

and $\displaystyle xy(x^2-y^2) = x^2 + y^2$ becomes $\displaystyle x^4z(1-z^2) = x^2(1+z^2)$

$\displaystyle \Rightarrow x^2 = \frac{1+z^2}{z(1-z^2)}, \, x\ne 0$

So we need the minimum value of $\displaystyle x^2(1+z^2)$ i.e. $\displaystyle \frac{(1+z^2)^2}{z(1-z^2)}$

Sorry, I've no more time at present to investigate further.

This does indeed give a solution. The value of the expression must be positive, which means $\displaystyle z < -1$ or $\displaystyle 0<z<1$. If you differentiate, and put the result equal to zero, you get a quadratic in $\displaystyle z^2$, which gives values of $\displaystyle z$ in the permissible ranges of
$\displaystyle z = \sqrt{3 - \sqrt8}= \sqrt2 -1$
and $\displaystyle z = -\sqrt{3+\sqrt8}= -\sqrt2 - 1$
Substituting either of these values back gives the minimum value of $\displaystyle x^2 +y^2$ as exactly $\displaystyle 4$, but there's a lot of manipulation of surds along the way. (I've checked this numerically on a spreadsheet, and am pretty sure this is correct.)