$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$ Prove the following inequality
$$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}, \enspace \forall a,b,c \in (0,\infty)$$
I tried by multying both sides by the denominator $(a+b+c)^2$ and then applying Holder for the left side but I couldn’t work it out. I  would prefer a proof without never ending computations (a.k.a. Brute Force / opening up the brackets) because I know how to do it this way already. A proof using C-B-S, Holder, Titu’s Lemma or their generalizations or other well-known inequalities would be ideal. Thank you!
 A: This seems to be deceptively cumbersome, needing a tight bound on the LHS while symmetrisation.  The following is from my old notes, unfortunately no way to attribute it correctly....  

First we need a well known inequality $4(x+y+z)^3 \geqslant 27(x^2y+y^2z+z^2x+xyz)$, which follows from AM-GM, as WLOG we may assume $y$ is in between $x, z$:
$$\frac{\frac{x+z}2+y+\frac{x+z}2}3\geqslant \sqrt[3]{\frac{x+z}2\cdot y \cdot \frac{x+z}2}$$
$$\implies \frac4{27}(x+y+z)^3 = x^2y+y^2z+z^2x+xyz - z(y-x)(y-z) \\\geqslant (x^2y+y^2z+z^2x+xyz)$$

Using the above with $x = \frac{a}b, y=\frac{b}c, z = \frac{c}a$, we get
$$\left(\frac{a}b+\frac{b}c+\frac{c}a \right)^3 \geqslant \frac{27}4\left( \frac{a^3+b^3+c^3}{abc}+1\right)$$
Hence to prove the original inequality, it is enough to show instead the symmetric
$$\frac{a^3+b^3+c^3}{abc} +1\geqslant 108 \frac{(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{1}$$
$$\iff \frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{abc}+4\geqslant \frac{108(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{2}$$
Using the obvious $(ab+bc+ca)^2 \geqslant 3abc(a+b+c)$,
$$\frac{a+b+c}{abc} \geqslant \frac{3(a+b+c)^2}{(ab+bc+ac)^2}$$
and hence, it is enough to show
$$\frac{3(a+b+c)^2(a^2+b^2+c^2-ab-bc-ac)}{(ab+bc+ac)^2} + 4 \geqslant \frac{108(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{3}$$
Setting 
$t=\dfrac{3(a^2+b^2+c^2)}{(a+b+c)^2}\, \in [1, 3)$
we need to equivalently show
$$\frac{54(t-1)}{(3-t)^2}+4\geqslant 4t^3 \tag{4}$$
which is $(t-1)(9-6t-8t^2+10t^3-2t^4)\geqslant 0$
and hence holds true as
$$9-6t-8t^2+10t^3-2t^4=2(3+3t-t^2)(t-1)^2+3>0$$
A: There is also the following proof (not mine) by SOS.
$$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a}-\frac{9(a^2+b^2+c^2)}{(a+b+c)^2}=\frac{\sum\limits_{cyc}\left((a^2-ab)abc+a(a-b)^2(b-2c)^2\right)}{abc(a+b+c)^2}\geq0.$$
A: There is also another following proof (not mine) by SOS, it is nice although I can't understand how to get it.
$$\text{LHS-RHS} = \frac{1}{16(a+b+c)^2}\sum \frac{[15c^2 +(4b-7c)^2](a-b)^2}{bc}$$
