are for every $\alpha \in N $ , $ \frac{1}{\alpha-2} \geq \frac{1}{\alpha}+\frac{1}{\alpha-1}$?
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1$\begingroup$ Maybe you mean reverse inequality? $\endgroup$ – zkutch Nov 26 '20 at 10:47
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$\begingroup$ 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
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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 $\alpha=3$.
