This is very similar to 2018 USAMO P1. Since everything is homogenous, wlog set $a=1$. Let $t=b+c$. We first prove that $t\ge 1$. By AM-GM, $$5bc=1+b^3+c^3=8\times \frac18 + b^3+c^3\ge 10\sqrt[10]{2^{-24}b^3c^3}\iff$$$$2^{-10}b^{10}c^{10}\ge 2^{-24}b^3c^3\iff bc \ge \frac14$$. Thus, by AM-GM again, $b+c\ge 2\sqrt{bc}\ge 1$. We can solve for $bc$ in terms of $t$: $$1+b^3+c^3=5bc\iff 1+t(t^2-3bc)=5bc\iff bc = \frac{t^3+1}{5+3t}$$. Our inequality rewrites, in terms of $t$, as $$(b+c)(bc+b+c+1)\ge 9bc\iff t(t+1)\ge (9-t)bc\iff t(3t+5)\ge (9-t)(t^2-t+1)\iff$$$$t^3-7t^2+15t-9\ge 0\iff$$$$(t-3)^2(t-1)\ge 0$$is true since $t\ge 1$ and the problem is solved. Note: I'm pretty sure $bc=\frac14$ is true as well, since we have $5bc=1+b^3+c^3\ge 1+(bc)^{\frac{3}{2}}\iff (2\sqrt{bc}-1)(\sqrt{bc}-1)^2\le 0\implies \sqrt{bc}\le \frac12$. Motivation comes from equality case at $a=1,b=c=0.5$.

DylanN wrote: Let $a, b, c$ be positive real numbers such that $a^3 + b^3 + c^3 = 5abc$. Show that \[ \left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \geq 9. \] Let $a+b+c=3u$, $ab+ac+bc=3v^2$ , $abc=w^3$ and $u^2=tv^2$. Thus, the condition gives $$w^3=\frac{27}{2}(u^3-uv^2)$$and since $$(a-b)^2(a-c)^2(b-c)^2\geq0,$$we obtain: $$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0$$or $$(15t-16)(63t^2-66t-1)\leq0,$$which gives $$\frac{11+8\sqrt2}{21}\leq t\leq\frac{16}{15}.$$Id est, $$\prod_{cyc}\frac{a+b}{c}=\frac{9uv^2-w^3}{w^3}=\frac{5v^2-3u^2}{3u^2-3v^2}=-1+\frac{2}{3(t-1)}\geq-1+\frac{2}{3\left(\frac{16}{15}-1\right)}=9.$$

DylanN wrote: Let $a, b, c$ be positive real numbers such that $a^3 + b^3 + c^3 = 5abc$. Show that \[A=\left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \geq 9.\] Let $x=\frac{a+b}{c}, y=\frac{b+c}{a}$ and $ z=\frac{c+a}{b} $. From $$ \sum \frac{a+b}{c}\cdot \frac{b+c}{a}= \frac{(a+b)(b+c)(c+a)}{abc}+\frac{a^3+b^3+c^3}{abc}+1$$and $$ \sum \frac{a}{b+c} \cdot \frac{b}{c+a} +\frac{2abc}{(a+b)(b+c)(c+a)}=1 $$we get $xy+yz+zx=xyz+6$ and $x+y+z=xyz-2$. So, $x,y,z$ are positive real roots of cubic polynomial: $$ t^3-(A-2)t^2+(A+6)t-A$$,which has discriminant $$ \Delta =-(A-9)(3A+5)(A^2-8A-16) \ge 0$$Hence, $ 9 \le A \le 4 + 4\sqrt{2}. $

Better: $a,b,c>0$ then $$ \frac{(a+b)(b+c)(c+a)}{abc} \ge \frac{11}{2} + \sqrt {3 \cdot \frac{a^3+b^3+c^3}{abc}- \frac{11}{4}} $$ Equality holds at $(a,b,c) ~ (1,1,1)$ or $(1,1,2) $

KaiRain wrote: Better: $a,b,c>0$ then $$ \frac{(a+b)(b+c)(c+a)}{abc} \ge \frac{11}{2} + \sqrt {3 \cdot \frac{a^3+b^3+c^3}{abc}- \frac{11}{4}} $$Equality holds at $(a,b,c) ~ (1,1,1)$ or $(1,1,2) $ MV help here.

KaiRain wrote: Better: $a,b,c>0$ then $$ \frac{(a+b)(b+c)(c+a)}{abc} \ge \frac{11}{2} + \sqrt {3 \cdot \frac{a^3+b^3+c^3}{abc}- \frac{11}{4}} $$ Equality holds at $(a,b,c) ~ (1,1,1)$ or $(1,1,2) $ $uvw$ kills it immediately because we need to prove that: $$18uv^2\geq13w^3+\sqrt{(324u^3-324uv^2)w^3+25w^6}.$$

DylanN wrote: Let $a, b, c$ be positive real numbers such that $a^3 + b^3 + c^3 = 5abc$. Show that \[ \left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \geq 9. \] From condition we get $a,\,b,\,c$ be the lengths of the sides of a triangle (see), therefore (see) \[2[ab(a+b)+bc(b+c)+ca(c+a)] \geqslant a^3+b^3+c^3+9abc.\]Then #1 is the consequence of \[ (a+b)(b+c)(c+a)-9abc + \frac{5abc-a^3-b^3-c^3}{2} = \frac{\displaystyle 2 \sum ab(a+b)-a^3-b^3-c^3-9abc}{2} \geqslant 0.\]

DylanN wrote: Let $a, b, c$ be positive real numbers such that $a^3 + b^3 + c^3 = 5abc$. Show that\[ \left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \geq 9.\] Let $x=\frac{a}{a+b+c}$, $y=\frac{b}{a+b+c}$, $z=\frac{c}{a+b+c}$, then $$x+y+z=1$$and $x^3+y^3+z^3=5xyz\Longleftrightarrow$ $$2xyz+3(xy+yz+zx)=1$$hence $$(3+2x)(3+2y)(3+2z)=49$$By AM-GM Inequality we have $$49\le (3+2x)\left(\frac{6+2(y+z)}{2}\right)^2=(3+2x)(4-x)^2\Longrightarrow 3-2\sqrt2\le x\le\frac12$$We need to prove that $$(1-x)(1-y)(1-z)\ge 9xyz\Longleftrightarrow xy+yz+zx\ge 10xyz\Longleftrightarrow 32xyz\le 1$$But $$32xyz-1=\frac{32x(3x^2-3x+1)}{2x+3}-1=\frac{3(2x-1)(4x-1)^2}{2x+3}\le0$$done.

The following statement is very nice. Let $a,b,c$ be positive reals and let $x,y$ such that: \begin{align*} x= \frac{a^3+b^3+c^3}{abc}-3 \end{align*}\begin{align*} y= \frac{\left(a+b\right) \left(b+c \right) \left(c+a \right) }{abc}-8\end{align*}Prove that $x-y \ge \left(\frac{2x}{y}-3\right)^2$.

Let $a, b, c$ be positive real numbers such that $a+b+c=5\sqrt[3]{abc}.$ Show that $$ \frac{81}{4} \le \left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \le 4 + 20\sqrt{2}$$https://artofproblemsolving.com/community/c5h1629600p23585834

We see that when $(a,b,c)=(1,1,2)$ then equality holds. This motivates to prove that $a,b,c$ are side lengths of a triangle (maybe degenerate). Let $b+c=x$. Then $\frac{5ax^2}{4}\ge 5abc=a^3+b^3+c^3\ge a^3+\frac{x^3}{4}=\frac{4a^3+x^3}{4}\ge\frac{5}{4}\sqrt[5]{a^{12}x^3}$ $\implies a^5x^{10}\ge a^{12}x^3\implies x^7\ge a^7\implies x\ge a$. We finish by the identity $$(a+b)(b+c)(c+a)-9abc=(a+b-c)(b+c-a)(c+a-b)+a^3+b^3+c^3-5abc.$$

Let $a=\max\{a,b,c\}$. We'll prove that $b+c\geq a$. Indeed, let $bc=v^2$, where $v>0$. Thus, $a\geq v$ and $$5abc=5av^2=a^3+b^3+c^3\geq a^3+2v^3,$$which gives $$a^3-5v^2a+2v^3\leq0$$or $$(a-2v)(a^2+2va-v^2)\leq0$$or $$a\leq2v,$$which gives $$a\leq2v=2\sqrt{bc}\leq b+c.$$

WLOG $c=1$ since the inequality and condition are homogenous. We need to show: $$(a+b)(a+1)(b+1)\ge9ab$$if $a^3+b^3+1=5ab$. Expanding, this is equivalent to: $$a^2b+ab^2+a^2+2ab+b^2+a+b\ge9ab.$$Let $t=a+b$, then $ab=\frac{t^3+1}{3t+5}$. Thus, we need to show: $$\frac{t^4+t}{3t+5}+t^2+t\ge\frac{9t^3+9}{3t+5}$$or $$(t+1)(t-1)(t-3)^2\ge0.$$Thus it suffices to show that $t\ge1$. But this is true since: $$\frac{t^3+1}{3t+5}=ab\le\left(\frac{a+b}2\right)^2=\frac{t^2}4$$$$\Leftrightarrow(t-1)(t^2-4t-4)\le0.$$If $t<1$ then $t^2-4t-4<0$, contradiction. Thus we are done.

Let $a,b,c$ be positive real numbers such that $a^3+b^3+c^3=5abc=5$. Show that$$(a+b+c)^3\geq 32.$$ is equivalent

Let $a, b, c$ be positive real numbers such that $a^3 + b^3 + c^3 \geq 5abc$. Show that $$(a+b+c)\left( \frac{1}{a} +\frac{1}{b}+ \frac{1}{c}\right)\geq 10$$

Let $a, b, c$ be positive real numbers such that $a^3 + b^3 + c^3 = 5abc .$ Show that $$9\leq \left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \leq 4(\sqrt 2+1) $$

asdf334 wrote: Let $a,b,c$ be positive real numbers such that $a^3+b^3+c^3=5abc=5$. Show that$$(a+b+c)^3\geq 32.$$is equivalent $5(a^2+b^2+c^2)\le 6(ab+bc+ca)$ is also equivalent

Let $a=\max\{a,b,c\}$. Another proof for $2\sqrt{bc}\geq a$: $$8\cdot 5abc=7a^3+(a^3+8b^3+8c^3)\ge 7a^3+12 abc$$$$ \Rightarrow \ \ 28abc\ge 7a^3 \ \ \ \Rightarrow \ \ 4bc\ge a^2$$

Let $a, b, c$ be positive real numbers such that $a^3 + b^2c + bc^2=5abc.$ Show that $$9\leq \left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \leq \frac{4}{27} (86+17\sqrt{34})$$Let $a, b, c$ be positive real numbers such that $a^2b + b^2c + c^2a=5abc.$ Show that $$\frac{45}{4} \leq \left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \leq 4(2+\sqrt{2})$$

Let $a, b, c$ be positive real numbers such that $(a+b+c)\left( \frac{1}{a} +\frac{1}{b}+ \frac{1}{c}\right)\leq 10.$ Prove that$$a^3 + b^3 + c^3 \le 5abc$$

DylanN wrote: Let $a, b, c$ be positive real numbers such that $a^3 + b^3 + c^3 = 5abc$. Show that \[ \left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \geq 9. \] sqing wrote: Let $a, b, c$ be positive real numbers such that $a^3 + b^3 + c^3 \geq 5abc$. Show that $$(a+b+c)\left( \frac{1}{a} +\frac{1}{b}+ \frac{1}{c}\right)\geq 10$$ $$\iff$$$$ \frac{a + b}{c}+\frac{b + c}{a} +\frac{c + a}{b} \geq 7$$anhduy98: $$(a+b+c)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=10+\frac{(a+b+c)(a+b-c)(a+b-3c)^2+(a^3+b^3+c^3-5abc)(9c-a-b)}{abc(3a+3b+5c)}.$$

Let $ a, b, c>0 $ and $a^3 + b^3 + c^3 = 5abc$. Show that $$9\leq \left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \leq 4(1+\sqrt 2)$$$$7 \leq \frac{a + b}{c}+\frac{b + c}{a} +\frac{c + a}{b} \leq 2+4\sqrt 2$$