If $a,b,c$ are postive real numbers. Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{16(ab+bc+ca)^2}\ge \frac{1}{a+b+c}.\left(\dfrac{1}{\sqrt{abc}}-3\right).$$ My trying based on a guess equality point $a=b=c=\dfrac{1}{\sqrt[3]{576}}.$ By multiply $a+b+c,$ we will prove $$\frac{(a+b+c)(ab+bc+ca)}{abc}+3+\frac{a+b+c}{16(ab+bc+ca)^2}\ge \dfrac{1}{\sqrt{abc}},$$ which I can't go futher.
I want to ask for help here. Thank you.
After assuming $a=b=c$ we obtain $a=\frac{1}{\sqrt[3]{576}}.$
From here let $a=\frac{x}{\sqrt[3]{576}}$, $b=\frac{y}{\sqrt[3]{576}}$and $c=\frac{z}{\sqrt[3]{576}}$.
Thus, we need to prove that: $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{36}{(xy+xz+yz)^2}\geq\frac{1}{x+y+z}\left(\frac{24}{\sqrt{xyz}}-3\right)$$ or $$(xy+xz+yz)(x+y+z)+3xyz+\frac{36xyz(x+y+z)}{(xy+xz+yz)^2}\geq24\sqrt{xyz},$$ which is true by AM-GM and C-S: $$(xy+xz+yz)(x+y+z)+3xyz+\frac{36xyz(x+y+z)}{(xy+xz+yz)^2}=$$ $$=\sum\limits_{cyc}(x^2y+x^2z+2xyz)+\frac{36xyz(x+y+z)}{(xy+xz+yz)^2}\geq$$ $$\geq2\sqrt{\frac{\sum\limits_{cyc}(x^2y+x^2z+2xyz)\cdot36xyz(x+y+z)}{(xy+xz+yz)^2}}=$$ $$=12\sqrt{\frac{\sum\limits_{cyc}x(y+z)^2\sum\limits_{cyc}x\cdot xyz}{(xy+xz+yz)^2}}\geq12\sqrt{\frac{\left(\sum\limits_{cyc}x(y+z)\right)^2\cdot xyz}{(xy+xz+yz)^2}}=24\sqrt{xyz}.$$