How to prove that $f$ is $1-1$ from $E$ on $\{ (s,t) : s> 2\sqrt{t} >0\}$

Question

Question:

Let $E=\{(x,y): 0<y<x \}$ set $f(x,y)=(x+y, xy)$ for $(x,y)\in E$

a) How to prove that $f$ is $1-1$ from $E$ on $\{ (s,t) : s> 2\sqrt{t} >0\}$

And how to find formula for $f^{-1}(s,t)$

b) compute $Df^{-1}(f(x,y))$ for $(x,y)\in E$ by using inverse function theorem.

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answer for part b)

I firstly compute $D(f(x,y))$

$$D(f(x,y))=\begin{pmatrix} 1 && 1 \\ y && x\end{pmatrix} $$

Then $$Df^{-1}=\begin {pmatrix} \frac{-x}{y-x} && \frac{1}{y-x} \\ \frac{y}{y-x} && \frac{-1}{y-x}\end{pmatrix}$$

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I did part-b. Hopefully it is true. Please check my answer. Also I could not do part-a. Please help me to solve this. Thank you.

Answer 1

HINT: For part $1$ $$f(x,y)=(x+y,xy)=(s,t)$$ hence $$s=x+y>2\sqrt{xy}=2\sqrt{t}$$ by AM-GM inequality since $x>0,y>0$.

and $$x+\frac{t}{x}=s\Rightarrow x=\frac{s\pm\sqrt{s^2-4t}}{2}$$ But $y<x$ constarint forces $$x=\frac{s+\sqrt{s^2-4t}}{2},\ y=\frac{s-\sqrt{s^2-4t}}{2}$$ hence $1-1$.

Answer 2

Hints for (a):

$$f(x,y)=f(a,b)\iff (x+y,xy)=(a+b,ab)\iff\begin{align*}x+y&=a+b\\xy&=ab\end{align*}$$

But then

$$b=\frac{xy}a\implies x+y=a+b=a+\frac{xy}a\implies a^2-(x+y)a+xy=0\implies$$

$$\Delta=(x+y)^2-4xy=(x-y)^2\left(\ge 0\right)\implies a_{1,2}=\frac{(x+y)\pm(x-y)}{2}=\begin{cases}x\\{}\\y\end{cases}$$

Take it now from here...