# any one can prove following inequality?

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

• Maybe you mean reverse inequality? – zkutch Nov 26 '20 at 10:47
• 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! – D.W. Nov 28 '20 at 22:17

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 $$\alpha=3$$.