Bump to this beauty

I claim the only such $r$ is $r=1$. Step 1: $r<0$ and $r>3$ are impossible. Note that if $r<0$ then by making $a,b,c$ arbitrarily large, we obtain a contradiction. Likewise, if $r>3$, then we obtain a similar contradiction by making $a,b,c$ arbitrarily small. Thus $0\le r\le 3$. Step 2. Impossibility of $0\le r\le 3$ except $r=1$. Now, see that $r=0,3$ does not work. Assume $r\in(0,3)$. Take $a,b,c$ such that $$a^3 = b^3 = c^3 = \frac{3-r}{2r}.$$ With this, $$r(ab+bc+ca) + (3-r)\left(\frac1a+\frac1b+\frac1c\right) = (ab+bc+ca)\left(r+\frac{3-r}{a^3}\right) = 9r\left(\frac{3-r}{2r}\right)^{\frac23}.$$ Now, we show this quantity is always below $9$. Assume the contrary and obtain $$r^3\frac{(3-r)^2}{4r^2} \ge 1 \iff r(3-r)^2 \ge 4 \iff (r-4)(r-1)^2 \ge 0,$$ implying $r\ge 4$ or $r=1$. But we already shown $r>3$ is impossible. Hence, the only remaining case is $r=1$. We claim $r=1$ works. To that end, it suffices to show $$(ab+bc+ca)\left(1+\frac{2}{abc}\right)\ge 9.$$ Set $abc=t^3$ and note by AM-GM that $ab+bc+ca\ge 3t^2$. With this, it suffices to have $$3t^2\left(1+\frac{2}{t^3}\right)\ge 9\iff \frac{(t-1)^2(t+2)}{t} \ge 0,$$ which is trivially true as $t>0$.

Marinchoo wrote: Find all real numbers $r$, such that the inequality $$r(ab+bc+ca)+(3-r)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9$$ holds for any real $a,b,c>0$. https://artofproblemsolving.com/community/c6h476867p24717131