Proving $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$, given $abc=1$ Let a, b, c be positive numbers such that $abc=1$. Prove that $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$
The usual methods do not seem to work, including a substitution $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ and trying to apply Muirhead's inequality.

$$\Longleftrightarrow (ab+bc+ac)+\dfrac{3(ab+bc+ac)}{a+b+c}\ge 6$$
since by $AM-GM$ inequality
$$ (ab+bc+ac)+\dfrac{3(ab+bc+ac)}{a+b+c}\ge 2\sqrt{\dfrac{3(ab+bc+ac)^2}{a+b+c}}$$
$$\Longleftrightarrow2\sqrt{\dfrac{3(ab+bc+ac)^2}{a+b+c}}\ge 6$$
$$\Longleftrightarrow (ab+bc+ac)^2\ge 3(a+b+c)$$
since by $AM-GM$
$$(ab+bc+ac)^2\ge 3abc(a+b+c)=3(a+b+c)$$
By Done.
