<s>[INST] Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$ for positive $a,b,c$ 
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.

Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$. 
 [/INST] Divide both sides of your inequality by (a+b+c), so you now have
$$bc+ac+ab\geq\frac {9abc}{(a+b+c)}$$
Now divide both sides by a so you get
$$\frac{(bc)}{a} + c +b \ge\frac {9bc}{(a+b+c)}$$
Next divide by b
$$\frac c a + \frac a b + 1 \ge \frac {9c}{(a+b+c)}$$
Finally, divide by c
$$\frac 1 a + \frac 1 b + \frac 1 c \ge \frac 9{(a+b+c)}$$
(I have used > instead of greater than or equal to because I can't get that symbol on my keyboard.)
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