In some derivation I arrived to I had reached the following assuming $\alpha,\beta >0$ $$\max\{1,1+\beta \}\leq \max\{\alpha,\alpha+\beta\}$$

Obviously the following condition can be deduced $$1\leq\alpha$$

Is there any other condition? Thanks

  • 2
    $\begingroup$ Notice that $\forall x\in \Bbb R\;\Big[\max\{x, x+\beta\} = x+\max\{0, \beta\}\Big]$ so the assumption $\alpha, \beta > 0$ is not even required. $\endgroup$ – Graham Kemp Aug 25 '15 at 0:50

$\max\{1,1+\beta\}=1+\beta$ and $\max\{\alpha,\alpha+\beta\}=\alpha+\beta$. Thus, the inequality you have reached is $$1+\beta\le\alpha+\beta$$ which is equivalent to $$1\le\alpha$$

  • $\begingroup$ I have edited my question... $\endgroup$ – Henry Aug 25 '15 at 0:45
  • $\begingroup$ I'm sorry. My former answer meant that some information about $\beta$ was needed, but I was wrong: the meaning of the inequality does not depend on $\beta$ (if $\beta$ were negative it would be easy to show the same conclusion: $1\le \alpha$). $\endgroup$ – ajotatxe Aug 25 '15 at 1:14

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