Comparing two variable expressions quickly

Question

Let us say $S_1=2xy^2+3xy$, and $S_2=3y^2+7x+7y+8$. Then, can we say that $S_1\ge S_2$ if $y\le x-2$ and $x,y\in\mathbb{N}$?

I think yes, but the usual quadratic function method is taking too much time, although I found that the discriminant of the final quadratic is $\ge0$ as I assume $x\ge5$. Actually, I am trying to prove $$4\left\lfloor\frac{xy}{2(y+1)}\right\rfloor\ge \left\lceil\frac{x}{y}\right\rceil\left(\frac{y-1}{2}\right)+\frac{x}{2}+y$$. Any other to prove this, or, is the inequality wrong? Any hints? Thanks beforehand.

Answer 1

can we say that $S_1\ge S_2$ if $y\le x-2$ and $x,y\in\mathbb{N}$?

No, we cannot.

If $y=1$, then $S_1-S_2=-2x-18\lt 0$ for $x\ge 3$.

If $y=2$, then $$S_1-S_2=7x-34\ \begin{cases}\lt 0&\text{for $x=4$}\\\\ \gt 0&\text{for $x\ge 5$}\end{cases}$$

If $y\ge 3$, then we have $2y^2+3y-7\gt 0$, so $$\begin{align}S_1-S_2&=x (2 y^2 + 3 y - 7) - 3 y^2 - 7 y - 8 \\\\&\ge (y+2)(2y^2+3y-7)-3y^2-7y-8 \\\\&=2\{(y^3-11)+2y(y-2)\}\gt 0\end{align}$$

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The inequality $$4\left\lfloor\frac{xy}{2(y+1)}\right\rfloor\ge \left\lceil\frac{x}{y}\right\rceil\left(\frac{y-1}{2}\right)+\frac{x}{2}+y$$ does not hold when $(x,y)=(5,2)$.

Answer 2

I will only answer the continous parts of this question.

In $\mathbb R^2$ neighbourhoods, calculate Hessian of $S_1-S_2$ in the points where $\nabla(S_1-S_2)$ is zero vector. Hessian needs to be definite pos/neg for local min/max.

On $\mathbb R^1$ neighbourhoods such as your line $y\leq x-2$ simplify function down to one variable and analyze it with ordinary 1 variable calculus.