$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc=\frac{3}{\sqrt[3]{abc}}-abc$ let $x=\sqrt[3]{abc},f(x)=\frac{9}{x}-x^3$ notice that $f(x)$ is decreasing in the interval $[0,1]$ $3=abc(a+b+c)\ge 3x^4\implies x\le 1$ so $f(x)\ge f(1)=8$

$$3=abc(a+b+c) \ge 3abc\sqrt[3]{abc}$$so $$abc\leq 1$$and: $$ LHS+abc=(a+b+c)(ab+ac+bc)$$so we have to prove: $$(a+b+c)(ab+ac+bc)\ge 8+abc$$because of $abc\leq1$ we know that $ 8+abc \leq 9$ so we just have to prove :$$(a+b+c)(ab+ac+bc)\ge 9$$but we know $$(ab+ac+bc)^2 \ge 3abc(a+b+c)=9$$so we have $ab+ac+bc\ge 3$ and we know $a+b+c\ge 3 $ problem solve

The sum can be done by $uvw$ method also....... But, the solution is very big and boring also

parsa1999 wrote: $$3=abc(a+b+c) \ge 3abc\sqrt[3]{abc}$$so $$abc\leq 1$$and: $$ LHS+abc=(a+b+c)(ab+ac+bc)$$so we have to prove: $$(a+b+c)(ab+ac+bc)\ge 8+abc$$because of $abc\leq1$ we know that $ 8+abc \leq 9$ so we just have to prove :$$(a+b+c)(ab+ac+bc)\ge 9$$but we know $$(ab+ac+bc)^2 \ge 3abc(a+b+c)=9$$so we have $ab+ac+bc\ge 3$ and we know $a+b+c\ge 3 $ problem solve why $a+b+c\ge 3$?

Reynan wrote: why $a+b+c\ge 3$? It's obvious \((a+b+c)^2\geq3(ab+bc+ca)\)

phymaths wrote: Let $a,b,c $ be real positive numbers such that $abc(a+b+c)=3$ Prove that $(a+b)(b+c)(c+a)> 8$ Simple! $\begin{array}{l} 3 = abc\left( {a + b + c} \right) \ge 3abc\sqrt[3]{{abc}} \Rightarrow abc \le 1 \Rightarrow a + b + c \ge 3\\ {\left( {ab + bc + ca} \right)^2} \ge 3abc\left( {a + b + c} \right) = 9 \Rightarrow ab + bc + ca \ge 3\\ \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) = \left( {a + b + c} \right)\left( {ab + bc + ca} \right) - abc \ge 9 - abc \ge 9 - 1 = 8. \end{array}$. Done!

phymaths wrote: Let $a,b,c $ be real positive numbers such that $abc(a+b+c)=3$ Prove that $(a+b)(b+c)(c+a)> 8$ JBMO 2010 Shortlist A4

Factorize this and use given facts, so: $(a+b)(b+c)(c+a)=(ab+ac+b^2+bc)(c+a)=(b(a+b+c)+ac)(c+a)=(\frac{3b}{abc}+ac)(c+a)=(\frac{3}{ac}+ac)(c+a)$. Now, from $AM-GM$ inequality $\frac{3}{ac}+ac=\frac{1}{ac}+\frac{1}{ac}+\frac{1}{ac}+ac \geq 4\sqrt[4]{\frac{ac}{a^3c^3}}$ and $c+a \geq 2\sqrt{ac}$. Finally: $(\frac{3}{ac}+ac)(c+a) \geq 4\sqrt[4]{\frac{ac}{a^3c^3}} \cdot 2\sqrt{ac}=8$. And, that is all

We have $3=abc(a+b+c)\geq abc(3\sqrt[3]{abc}) \implies 1\geq \sqrt[3]{abc}$. Rewriting the given inequality as \[3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 8+abc\]so it is suffices to prove that \[\frac{9}{\sqrt[3]{abc}}\geq 8+abc\]letting $t=\sqrt[3]{abc}$ gives \[(t-1)(t^3+t^2+t+9)\leq 0\iff t\leq 1\]which is true.

Probably similar to 2above, but I found it a long time ago and thought it was pretty interesting: We have $$(a+b)(b+c)(c+a)= ab(a+b+c)+bc(a+b+c)+ac(a+c)=\frac{3}{c}+\frac{3}{a}+a^2c+ac^2\ge 8\sqrt[8]{\frac{a^3c^3}{a^3c^3}}=8$$

Notice that, $$(a+b)(b+c)(c+a)=ab(a+b+c)+bc(a+b+c)+ac(a+c)=\frac{3}{c}+\frac{3}{a}+ac(a+c)$$By AM-GM, we have $$\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{ac(a+c)}{2}+\frac{ac(a+c)}{2}\ge8\sqrt[8]{\frac{(a+c)^2}{4ac}}\ge8$$Hence, proved.

The condition is that $(ab)(bc)+(bc)(ca)+(ca)(ab)=3$. From Maclaurin, $ab+bc+ca \geq 3$. From Maclaurin again, $a+b+c \geq 3$, so $(a+b+c)(ab+bc+ca) \geq 9$. From the 8-9 inequality, $(a+b)(b+c)(c+a) \geq \frac{8}{9}(a+b+c)(ab+bc+ca) = 8$.

$3=abc(a+b+c) \geq 3abc \sqrt[3]{abc} \implies 1 \geq abc$ $\Pi{(a+b)}=\Pi{(\frac{3}{abc}-c)=\Pi(\frac{3-abc^2}{abc}}) \geq 8$ $\Pi{(3-a^2bc)} \geq 8a^3b^3c^3$ $LHS=27-9\sum{a^2bc}+3\sum{ab^3c^3}-a^4b^4c^4 \geq 8a^3b^3c^3$ $\sum{a^2bc}=27$ so $3\sum{a^2b^2} \geq a^3b^3c^3+8a^2b^2c^2$ We have \[\frac{a^2b^2+b^2c^2+c^2a^2}{3} \geq \sqrt[3]{a^4b^4c^4} \geq a^3b^3c^3\]\[\frac{8(a^2b^2+b^2c^2+c^2a^2)}{3} \geq 8\sqrt[3]{a^4b^4c^4} \geq 8a^2b^2c^2\]By adding these two inequalities, we get the desired inequality.