MHB An upperbound for an algebraic expression

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The discussion revolves around finding a tight upper bound for the expression x/(1-x(1-y)) with constraints 0 ≤ x, y ≤ 1. Participants conclude that there is no upper or lower bound for the function due to its behavior on critical curves. Specifically, as y approaches 0 and x approaches 1, the function tends to infinity. Additionally, on the critical curve x = 1/(1-y), the function can also reach negative infinity. Thus, the consensus is that the function does not have a defined upper bound.
bincy
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Hii Everyone,
Plz tell me a tight upperbound for [math]\frac{x}{1-x\left(1-y\right)}
[/math] where o<=x,y<=1

regards,
Bincy
 
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I don't think there is an upper or lower bound for that function. On the critical curve $x=1/(1-y)$, the function goes to negative infinity, and as, say, $y\to 1/2$ and $x\to 2$, the function goes to positive infinity.
 
Ackbach said:
I don't think there is an upper or lower bound for that function. On the critical curve $x=1/(1-y)$, the function goes to negative infinity, and as, say, $y\to 1/2$ and $x\to 2$, the function goes to positive infinity.
Plz note that BOTH x and y are in between 0 and 1
 
Ah, my mistake. You still have a problem when $y\to 0$ and $x\to 1$. The function blows up there, and there is no upper bound on it.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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