# Comparing two variable expressions quickly

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.

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

• 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 $x\ge y+2, x,y\in\mathbb{N}$ – vidyarthi May 11 at 14:26
• @vidyarthi: No, we cannot. Take $(x,y)=(11,7)$. There are many other counterexamples. – mathlove May 11 at 17:18

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.

• But your calculation would provide the local extrema, whereas I am just interested in proving whether the inequality holds. How could your method work – vidyarthi May 9 at 11:02
• @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 $S_2-S_1 \leq 0 \Leftrightarrow S_2\leq S_1$ – mathreadler May 9 at 11:06
• 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. $S_1\ge S_2$? – vidyarthi May 9 at 11:09
• 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. – mathreadler May 9 at 11:21