devious_ wrote:

No. Same mistake.

\[ abc(a + b + c) \geqslant bc + ca + ab \geqslant \sqrt {3abc(a + b + c)} \] so, \[ abc(a + b + c) \geqslant 3 \]

siuhochung, can you please explain how you arrived at $\sqrt {3abc(a + b + c)}$?

devious_ wrote: siuhochung, can you please explain how you arrived at $\sqrt {3abc(a + b + c)}$? It is just $(x+y+z)^2\geq 3(xy+yz+zx)$.

I was just coming to edit my post and say that I figured it out, then I noticed two replies.

I don't know whether it has a name, but you see that it is $x^2+y^2+z^2\geq xy+yz+zx$, which is very well known and widely used.

Its most ancient ancestor is called the trivial inequality.

Valentin Vornicu wrote: Let $a,b,c$ be positive numbers such that $a+b+c \geq \dfrac 1a + \dfrac 1b + \dfrac 1c$. Prove that \[ a+b+c \geq \frac 3{abc}. \] easy! my solution = the given condition is $abc(a+b+c)\ge ab+bc+ca$ but $(ab+bc+ca)^2\ge 3abc(a+b+c)\ge 3(ab+bc+ca)\Longrightarrow ab+bc+ca\ge 3\Longrightarrow abc(a+b+c)\ge 3$ so we are done

Let $a,b,c$ be positive real numbers such that: $abc(a+b+c) =3.$ Prove that $$a+b+c \geq \frac{3}{ab+bc+ca}+\frac{2}{abc}$$$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{3}{a+b+c}+\frac{2}{abc}$$Let $a,b,c$ be positive real numbers, such that: $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$ Prove that $$a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}$$https://artofproblemsolving.com/community/c6h1490860p8749593 Valentin Vornicu wrote: Let $a,b,c$ be positive numbers such that $a+b+c \geq \dfrac 1a + \dfrac 1b + \dfrac 1c$. Prove that \[ a+b+c \geq \frac 3{abc}. \] https://artofproblemsolving.com/community/c6h1184601p5748463