For $a, b, c$ positive reals prove that $$(ab+bc+ca)\bigg(\frac{a}{b(a^2+2b^2)}+\frac{b}{c(b^2+2c^2)}+\frac{c}{a(c^2+2a^2)}\bigg) \ge 3$$
I used the Cauchy-Swartz inequality in the LHS so I was left to prove that $\dfrac{a}{\sqrt{a^2+2b^2}}+\dfrac{b}{\sqrt{b^2+2c^2}}+\dfrac{c}{\sqrt{c^2+2a^2}} \ge \sqrt3$ but apparently for $(a,b,c)=(1,2,1)$ it is not correct.
Let $a=\frac{1}{x}$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$.
Hence, by Holder $(ab+ac+bc)\sum\limits_{cyc}\frac{c}{a(c^2+2a^2)}=(x+y+z)\sum\limits_{cyc}\frac{x^2}{y(x^2+2z^2)}=$
$=(x+y+z)\sum\limits_{cyc}\frac{x^3}{xy(x^2+2z^2)}\geq\frac{(x+y+z)^4}{3\sum\limits_{cyc}(x^3y+2x^2yz)}$.
Thus, it remains to prove that $(x+y+z)^4\geq9\sum\limits_{cyc}(x^3y+2x^2yz)$ or
$\sum\limits_{cyc}(x^4-5x^3y+4x^3z+6x^2y^2-6x^2yz)\geq0$ or
$\sum\limits_{cyc}(2x^4-x^3y-x^3z+12x^2y^2-12x^2yz)\geq9\sum\limits_{cyc}(x^3y-x^3z)$ or $\sum\limits_{cyc}(x-y)^2(x^2+xy+y^2+6z^2)\geq9(x+y+z)(x-y)(x-z)(y-z)$.
Hence, it remains to prove the last inequality for $x\geq y\geq z$. But $(x+y+z)(x-y)(x-z)(y-z)\leq(x+y)xy\Leftrightarrow z(x^2+y^2+xz-z^2)\geq0$, which is obvious.
Id est, it's enough to prove $(x+y+z)^4\geq9\sum\limits_{cyc}(x^3y+2x^2yz)$ for $z\rightarrow0$.
Let $x=ty$. Hence, we need to prove that $(t+1)^4\geq9t^3$, which is AM-GM: $(t+1)^4=\left(3\cdot\frac{t}{3}+1\right)^4\geq\left(4\sqrt[4]{\frac{t^3}{27}}\right)^4>9t^3$.
Done!