minimum of $x^2+y^2$

Question

Find minimum of the expression $$x^2+y^2$$ knowing that $x, y, a$ are real numbers ( $a$ fixed real number) so $$x^2-y^2+2xy=a.$$

My solution is: $$ x^2-y^2+2xy=a <=>(x+y)^2-2y^2=a<=>(x+y+y\sqrt{2})(x+y-y\sqrt{2})=a.$$

I wrote for $a$ nonzero :

$$x+y+y\sqrt{2}=t, x+y-y\sqrt{2}=\frac{a}{t}.$$

Solving this system we obtain:

$$x=\frac{(\sqrt{2}-1)t^2+(\sqrt{2}+1)a}{2t\sqrt{2}}, y= \frac{t^2-a}{2t\sqrt{2}}.$$

We calculated:

$$x^2+y^2=(\frac{(\sqrt{2}-1)t^2+(\sqrt{2}+1)a}{2t\sqrt{2}})^2+(\frac{t^2-a}{2t\sqrt{2}})^2=\cdots=\frac{(2-\sqrt{2})t^2}{4}+\frac{(2-\sqrt{2})a^2}{4t^2}\geq2\sqrt{\frac{4-2}{16}\cdot t^2\cdot\frac{a^2}{t^2}} =\frac{|a|}{\sqrt{2}}.$$

Now I wonder if I thought well, if the result is correct.

What do you think?

Thank you very much!

Answer 1

By Lagrange Multipliers : define

$$h(x,y,\lambda):=x^2+y^2+\lambda(x^2-y^2+2xy-a)$$

$$\begin{align*}h'_x&=2x+\lambda(2x+2y)=0\iff \lambda=-1-\frac yx\\ h'_y&=2y+\lambda(-2y+2x)=0\iff\lambda=1-\frac xy\\ h'_\lambda&=x^2-y^2+2xy-a=0\end{align*}$$

From the first two relations above we get

$$2=\frac xy-\frac yx=\frac{x^2-y^2}{xy}\iff x^2-y^2-2xy=0$$

and then $\;4xy=a\;$ ...and etc.