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Let $x_{1},x_{2},y_{1}$ and $y_{2}$ be distinct positive real numbers. I would like to upper bound the following quantity: \begin{equation} \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. \end{equation} 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.

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  • $\begingroup$ Do you want a strict upper bound? If not, then $\log x \lt x$ would help? $\endgroup$ Commented Jan 14, 2020 at 7:59
  • $\begingroup$ The given expression tends to $\infty$ as $x_2 \to \infty$ and the bound you are trying to get tends to $0$. $\endgroup$ Commented Jan 14, 2020 at 8:01
  • $\begingroup$ @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 $\endgroup$
    – nemo
    Commented Jan 14, 2020 at 8:15
  • $\begingroup$ @DhanviSreenivasan I tried with $\log x \le x-1$, but couldn't get anything useful $\endgroup$
    – nemo
    Commented Jan 14, 2020 at 8:16
  • $\begingroup$ @nemo In that case you should delete the term 'absolute constant'. $\endgroup$ Commented Jan 14, 2020 at 8:19

1 Answer 1

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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.

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  • $\begingroup$ The $x_i$ and $y_i$ are distinct, I have edited the question. $\endgroup$
    – nemo
    Commented Jan 14, 2020 at 9:40

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