Prove that the unique zeros of $f(x,y)=a x +(1-a)y+xy$ when $x,y\in[0,1]$, is $x=y=0$.

Question

Prove that the unique zeros of the two-variables function: $$f(x,y)=a x +(1-a)y+xy$$ when $x,y\in[0,1]$, is $x=y=0$. Here, $a$ is a parameter between 0 and 1. I have no idea where to start. Any suggestions please?

Answer 1

Beside the good comments you already received from David C. Ullrich and A.G., you could also look at the extremum of function $$f=a x +(1-a)y+xy$$ Computing the derivatives $$f'_x=a+y$$ $$f'_y=1-a+x$$ So, the extremum is obtained for $x=a-1$, $y=-a$ and, for these values, $$f=-a(1-a)$$ which is negative if $0<a<1$. So, there is no possible solution except if each term is zero as A.G. answered. In fact $x$ and $y$ do not need to be bounded.

Answer 2

$ax + (1-a)y + xy = 0$

$(x+1-a)(a+y) = a(1-a)$

$\left(1 + \dfrac{x}{1-a}\right) \left( 1 + \dfrac ya \right) = 1$

since $1 + \dfrac{x}{1-a} \ge 1$ and $1 + \dfrac ya \ge 1$,

we must have $x = y = 0$