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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.

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


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 $(x,y)=(5,2)$.

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  • $\begingroup$ Can we prove the second inequality $4\left\lfloor\frac{xy}{2(y+1)}\right\rfloor\ge\left\lceil\frac{x}{y}\right\rceil\left(\frac{y-1}{2}\right)+\frac{x}{2}+y$ for all $x,y\neq(5,2)$ and $x\ge y+2, x,y\in\mathbb{N}$ $\endgroup$ – vidyarthi May 11 at 14:26
  • $\begingroup$ @vidyarthi: No, we cannot. Take $(x,y)=(11,7)$. There are many other counterexamples. $\endgroup$ – mathlove May 11 at 17:18
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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.

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  • $\begingroup$ But your calculation would provide the local extrema, whereas I am just interested in proving whether the inequality holds. How could your method work $\endgroup$ – vidyarthi May 9 at 11:02
  • $\begingroup$ @vidyarthi The function is smooth. Any global extrema (if they exist, i.e. the function is not unbounded) will be local extrema. So if you find extrema with values $\leq 0, \geq 0$ respectively of $S_2-S_1$, you can investigate if it will hold. Because $S_2-S_1 \leq 0 \Leftrightarrow S_2\leq S_1$ $\endgroup$ – mathreadler May 9 at 11:06
  • $\begingroup$ So you mean to say that if $S_1-S_2$ has a local minima at $0$ then the inequality is true, right, i.e. $S_1\ge S_2$? $\endgroup$ – vidyarthi May 9 at 11:09
  • $\begingroup$ As long as it does not have any local minima smaller than 0 and we know it is bounded below by 0 in any point not an extrema. $\endgroup$ – mathreadler May 9 at 11:21

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