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Let $t_1,\dots,t_m \ge 0$, and $\lambda_1, \dots, \lambda_m \ge 0$. Suppose \begin{equation} \begin{aligned} & t_1 + \dots + t_m = m(\lambda_1 + \dots + \lambda_m), \\ & t_1 \times \dots \times t_m \leq \lambda_1 \times \dots \times \lambda_m. \end{aligned} \end{equation} Can we prove the following inequality: \begin{equation} \sum_{i=1}^m \frac{1}{t_i +1} \ge \sum_{i=1}^m \frac{1}{\lambda_i +1}. \end{equation} I am stuck for a long time and have tried many methods, but have no idea at all. Can anyone prove it or provide a counterexample? Any help will be appreciated.

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  • $\begingroup$ How is this linear algebra? $\endgroup$ – Arturo Magidin Mar 18 at 3:17
  • $\begingroup$ fine, I edited it. $\endgroup$ – Ze-Nan Li Mar 18 at 3:23
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    $\begingroup$ the thing is close to majorization theory related to linear algebra in a way. $\endgroup$ – Toni Mhax Mar 18 at 3:50
  • $\begingroup$ @ToniMhax Thanks for the hint ! $\endgroup$ – Ze-Nan Li Mar 18 at 3:51
  • $\begingroup$ maybe $t_1+t_2=\lambda_1+\lambda_2$ yields the inequality when $m=2$. etc $\endgroup$ – Toni Mhax Mar 18 at 4:16
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A preliminary calculation gives the following counterexample, $$m=2,t_1=\frac{3+\sqrt 7}{2}, t_2=\frac{3-\sqrt 7}{2}, \lambda_1=1,\lambda_2=\frac{1}{2}.$$ You can double check if it is indeed a counterexample or there is a mistake in my calculation.

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  • $\begingroup$ It seems correct, but I am very curious that how did you find this counterexample? Anyway, thank you very much. $\endgroup$ – Ze-Nan Li Mar 18 at 3:43
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    $\begingroup$ Actually, it should not be hard to check that under your conditions, the inequality is true if and only if $\lambda_1 \lambda_2\ge 1$. From this, it should be easy to get a counterexample. $\endgroup$ – QZ0 Mar 18 at 3:50
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The sum constraint may be the wrong one if $t_1=4$ and $t_2=0$, $\lambda_1=1.9$, $\lambda_2=0.1$, $\begin{equation} \sum_{i=1}^2 \frac{1}{t_i +1}=1.2 \le \sum_{i=1}^2 \frac{1}{\lambda_i +1}\approx 1.25 \end{equation}$

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