0
$\begingroup$

are for every $\alpha \in N $ , $ \frac{1}{\alpha-2} \geq \frac{1}{\alpha}+\frac{1}{\alpha-1}$?

$\endgroup$
  • 1
    $\begingroup$ Maybe you mean reverse inequality? $\endgroup$ – zkutch Nov 26 '20 at 10:47
  • $\begingroup$ The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! $\endgroup$ – D.W. Nov 28 '20 at 22:17
3
$\begingroup$

That inequality is not defined for $\alpha = 0, 1, 2$. For $\alpha > 2$ you have:

$$ \frac{1}{\alpha-2} - \frac{1}{\alpha-1} - \frac{1}{\alpha} = \frac{\alpha(\alpha-1)-\alpha(\alpha-2)-(\alpha-1)(\alpha-2)}{\alpha(\alpha-1)(\alpha-2)} $$

Clearly the denominator is always positive. Regarding the numerator: $$ \alpha(\alpha-1)-\alpha(\alpha-2)-(\alpha-1)(\alpha-2) = \alpha(\alpha-1-\alpha+2) - (\alpha-1)(\alpha-2) =\\ \alpha - (\alpha-1)(\alpha-2) = \alpha - \alpha^2 + 2\alpha + \alpha -2= -\alpha^2 + 4\alpha - 2, $$ which is negative as soon as $\alpha \ge 4$. In fact, your inequality is only true for $\alpha=3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.