If I can rewrite an inequality $f(x,y)> g(x,y)$ as $ h(x,y)>k(x,y)$, does that mean $f(x,y)>g(x,y)$ If and only if $h(x,y)>k(x,y)$?
My reasoning/difficulty:
Suppose I can rewrite $f(x,y)>g(x,y)$ as $h(x,y)>k(x,y)$. Then $f(x,y)>g(x,y) \implies h(x,y) > k(x,y)$ (i.e. $h(x,y)>k(x,y)$ is necessary for $f(x,y)>g(x,y)$).
Then, if i reverse the steps I should be able to show the opposite direction -- that $h(x,y)>k(x,y)$ is sufficient for $f(x,y)>g(x,y)$)
However, is it always possible to reverse the steps? Intuitively I feel like it should be.
Also, what method could be used to prove such a statement (it seems to me like it would be a proof showing that any possible transformation of the original inequality is reversible, but how can I prove that any transformation is reversible? It seems impossible to be sure there is not some transformation being overlooked?)
Edit:
- I have tried to elaborate on what I mean by "re-write" in the comments.
- I think for the result to have a shot at being true we need to assume that $f(x,y)\neq 0$ and $g(x,y)\neq 0$ (and maybe also for $h(x,y)$ and $k(x,y)$?). Thanks for the answer that points this out.
Edit 2: Perhaps a simple and more concrete version of the question:
Consider the inequality $2x^2y + 3y^2 > xy +5$ and suppose that it holds.
This inequality implies (assuming $y\neq 0$) that $2x^2 - x > -3y + > \frac{5}{y}$
Therefore, $2x^2 - x > -3y + \frac{5}{y}$ if $2x^2y + 3y^2 > xy +5$
Restated question: Since I obtained $2x^2y + 3y^2 > xy +5$ from $2x^2 - x > -3y + \frac{5}{y}$, can i automatically say that $2x^2 - x > -3y + \frac{5}{y}$ only if $2x^2y + 3y^2 > xy +5$,
The rewritten inequality is obtained from the initial by elementary steps, “using operations like addition, subtraction, multiplication, division, and where we are not multiplying/dividing by 0, and if we multiply/divide by a negative number we change the sign of the inequality”. So we have to check when such an operation transforms an inequality to an equivalent inequality. This can be understood studying basic properties of inequalities. Below I briefly list some elementary equivalent transformations of inequalities.
$f<g$ iff $f+t<g+t$ for each real $t$ and each real $f$ and $g$.
$f<g$ iff $tf<tg$ for each positive real $t$ and each real $f$ and $g$.
$f<g$ iff $tf>tg$ for each negative real $t$ and each real $f$ and $g$.
$f<g$ iff $\exp(f)<\exp(g)$ for each real $f$ and $g$.
$f<g$ iff $\ln(f)<\ln(g)$ for each positive real $f$ and $g$.