Hello, if \( a^2+b^2=1+\frac{2ab}{a-b} \) then:
\( (a-b-1)[(a-b)(a-b+1)+2ab]=0 \), then \( a-b=1 \) or \( ab=\frac{(a-b)(a-b+1)}{-2} \),
- If \( ab=\frac{(a-b)(a-b+1)}{-2} \) and \( \sqrt{a-b}=a^2+5b \)
How \( \sqrt{a-b}=a^2+5b\implies a-b\geq0 \) then \( ab=\frac{(a-b)(a-b+1)}{-2}\leq0 \) - If \( ab=\frac{(a-b)(a-b+1)}{-2} \) and \( a-b=1 \)
Then \( ab=\frac{(1)(1+1)}{-2}=-1 \) - If \( a-b=1 \) and \( \sqrt{a-b}=a^2+5b \)
Then \( 1=a^2+5b \) and \( a-b=1 \)
the solutions in this case are \( (a,b)=(1,0) \) or \( (a,b)=(-6,-7) \)
and \( ab=1\cdot0=0 \) or \( ab=-6\cdot(-7)=42 \)
Therefore maximum of \( ab \) is \( 42 \)