A question on a proof of $y(y+1) \le (x+1)^2 \implies y(y-1) \le x^2$

I was reading the question posted here:Does $y(y+1) \leq (x+1)^2$ imply $y(y-1) \leq x^2$?. The solution posted was:

"Given $y^2+y≤x^2+2x+1$, if possible, let $x2<y(y−1)$. Clearly $y>1.$

Then $x^2+(2x+1)<y^2−y+(2x+1)$ So $y^2+y<y^2−y+2x+1$, which resolves to $y<x+1/2.$

Hence we also have $y−1<x−1/2.$ As $y>1$, the LHS is positive, and we can multiply the last two to get $y(y−1)<x^2−1/4⟹y(y−1)<x^2$, a contradiction." However, I don't think this is right because he had $y<x+1/2$, and in the last step he multiplied $y−1$ by $y$ and $x−1/2 by x+1/2$. However, this inequality is not in the same proportion anymore, because y does not equal x+1/2. It seems like he increased the value of the right side compared to the left, and then concluding that the right side is bigger/ It's like if I try to prove that $a \lt b$, and to solve this I say $a \lt b+5$. While this is true, this changes the inequality.

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Yes, it is right and a nice one, moreover.

He had $y \lt x + \frac{1}{2}$

which implies $y-1 \lt x - \frac{1}{2}$

He then multiplied them. He was even careful to mention that the terms were positive (which was correct, based on the assumption he started out with: $y(y-1) \gt x^2$ and combined with the assumption in the problem statement, $y \gt 0$).

I don't understand what you mean by: "not even in the same proportion anymore".

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Well yea but if you wanna solve the original inequality aren't you limited to just multiplying both sides by the same thing? That way the RHS is increased in proportion to the LHS. However, what he multiplied the RHS by is larger than what he multiplied the LHS by. –  Ovi Apr 21 '13 at 14:43
@Ovi: There is no "solving" original inequality. One wants to show a statement about $x,y$ implies another statement. The proof is by contradiction. btw, Why does it have to be exact? Say, I want to prove that $2^{10} \lt 1600$. I can always prove it this way: $2^{5} \lt 40$, squaring gives $2^{10} \lt 1600$. There is nothing wrong there. –  Aryabhata Apr 21 '13 at 14:45
O oh I see. Thanks –  Ovi Apr 21 '13 at 14:49

As mentioned, $$0<y-1<x-\frac12,$$ and $$0<y<x+\frac12.$$ Hence, $$y(y-1)<y\left(x-\frac12\right)<\left(x+\frac12\right)\left(x-\frac12\right)=x^2-\frac14<x^2.$$

More generally, if $a<b$ and $c<d$, and we happen to know that $a,c>0,$ then we can always conclude that $ac<bd$.

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Yea I understand that but if you didn't know that a<b, could you use as an argument that ac<bd to prove that a<b? –  Ovi Apr 21 '13 at 14:46
Not in general, but that's not remotely what was done, here. It was simply a chain of inequalities, using ordered field properties of the reals, to induce a contradiction. That's a fairly standard technique of analysis. –  Cameron Buie Apr 21 '13 at 14:51
Oh ok I see it now haha I can't believe how I thought this was wrong. Thanks –  Ovi Apr 21 '13 at 14:56
To reword the proof given there: We start with real numbers $x,y$ such that $$\tag1 y\ge0$$ $$\tag2 y^2+y\le x^2+2x+1$$ $$\tag3 x^2<y(y-1)$$ If we had $y\le1$, then the right hand side in $(3)$ would be the product of a nonnegative (according to $(1)$) and a nonpositve (by assumtion) factor, hence nonnegative, contradicting $x^2\ge0$. Therefore $$\tag4 y>1.$$ Adding $2x+1$ to $(3)$ and combining with $(2)$ we find $y^2+y\le x^2+2x+1<y^2-y+2x+1$, i.e. $$\tag5 y\le x+\frac12.$$ By subtracting $1$, this becomes $$\tag6 y-1\le x-\frac12.$$ Because of $(4)$ and $(6)$, the number $x-\frac12$ is positive, hence we are allowed to multiply $(5)$ with $x-\frac12$. Also, we can multiply $(6)$ with the (again by $(4)$) positive number $y$ and combine this to find $$x^2\stackrel{(3)}<y(y-1)\stackrel{y\cdot(6)}\le y\left(x-\frac12\right)\stackrel{(x-\frac12)\cdot(5)}\le \left(x+\frac12\right)\left(x-\frac12\right)=x^2-\frac14<x^2,$$ contradiction.