Please help me for prove this inequality: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) &gt; 0$$
Given that $a,b,c$ are distinct positive real numbers, prove that $(a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)&gt;9$   This is how I tried doing it:  Let $p= a + b + c,$ and $q=\frac1{a}+ \frac1{b} + \frac1{c}$.  Using AM>GM for $p, q$, I get:  $$\frac{p+q}{2} &gt; {(pq)}^{1/2}$$ $$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} &lt; \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$ And for any $x\in \mathbb{R}, \space \space x+\frac1{x}≥2.$ Thus,   $$(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)&lt;9, $$ which is the opposite of what had to be proven. What did I do wrong?
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