Knowing $a,b,c>0$ and $abc\le1$, prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1+\frac{6}{a+b+c}$

Question

Knowing $a,b,c>0$ and $abc\le1$, prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1+\frac{6}{a+b+c}$

I tried to AM-GM the inequality, which gave this:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac}{1}{abc}\ge3$$
This, however, didn't lead me anywhere. Probably, I had to take Tittu of this, but:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{(1+1+1)^2}{a+b+c}=\frac{9}{a+b+c}$$
This didn't bring me anywhere, either. Any other help is welcomed.

Answer 1

I assume $a,\ b,\ c>0$.

For $a,\ b,\ c>0$ and $abc\leq1$, $\frac1{abc}\geq1$ and $\sqrt[3]{\frac{1}{abc}}\geq1$.

Also note that $a>b \land\ c>d\implies(a+c)>(b+d)$.

Now, by the AM-GM inequality:

$$\frac1a+\frac1b+\frac1c\geq3\sqrt[3]{\frac{1}{abc}}=\frac{2}{\sqrt[3]{abc}}+\sqrt[3]{\frac{1}{abc}}\geq\frac{2\cdot3}{a+b+c}+1=1+\frac6{a+b+c}.$$