# An upper bound on $\log\frac{x_{2}y_{2}}{x_{1}y_{1}}+\frac{1}{2}(x_{1}+y_{1})\bigg(\frac{1}{x_{2}}+\frac{1}{y_{2}}\bigg)-2$

Let $$x_{1},x_{2},y_{1}$$ and $$y_{2}$$ be distinct positive real numbers. I would like to upper bound the following quantity: $$$$\log\frac{x_{2}y_{2}}{x_{1}y_{1}}+\frac{1}{2}(x_{1}+y_{1})\bigg(\frac{1}{x_{2}}+\frac{1}{y_{2}}\bigg)-2.$$$$ Basically, I would like to get rid of the log term and get an overall $$\frac{f(x_{1},y_{1},x_{2},y_{2})}{x_{1}y_{1}x_{2}y_{2}}$$ type dependence if possible, where $$f$$ is allowed to depend on the differences $$(x_{1}-x_{2})$$ and $$(y_{1}-y_{2})$$.

Any suggestions appreciated.

EDIT: The problem has some more structure that I had abstracted out as I thought it wouldn't be required, but looks like as stated, the expression cannot be bounded above.

Each $$x_i$$ and $$y_i$$ is a sum of $$n$$ positive numbers and I am trying to get a $$1/n^4$$ dependence. The differences $$x_1-x_2$$ and $$y_1-y_2$$ can be bounded above by constants.

• Do you want a strict upper bound? If not, then $\log x \lt x$ would help? Commented Jan 14, 2020 at 7:59
• The given expression tends to $\infty$ as $x_2 \to \infty$ and the bound you are trying to get tends to $0$. Commented Jan 14, 2020 at 8:01
• @KaviRamaMurthy Right. Maybe $c$ can be allowed to have some dependence on $x_{i}$ and $y_{i}$, say their difference, which I can then assume to be upper bounded by some constant
– nemo
Commented Jan 14, 2020 at 8:15
• @DhanviSreenivasan I tried with $\log x \le x-1$, but couldn't get anything useful
– nemo
Commented Jan 14, 2020 at 8:16
• @nemo In that case you should delete the term 'absolute constant'. Commented Jan 14, 2020 at 8:19

Let $$x_1=y_1=kx_2=ky_2.$$
Thus, for $$k\rightarrow+\infty$$ we have:$$\log\frac{x_{2}y_{2}}{x_{1}y_{1}}+\frac{1}{2}(x_{1}+y_{1})\bigg(\frac{1}{x_{2}}+\frac{1}{y_{2}}\bigg)-2=2(k-\log{k}-1)\rightarrow+\infty,$$ which says that a maximal value does not exist.
• The $x_i$ and $y_i$ are distinct, I have edited the question.