$$8(a+b+c)^2 \overset{?}{\geq} 9\sqrt[3]{(1+(a+b)^2)(1+(b+c)^2)(1+(c+a)^2)}$$By AM-GM, \[9\sqrt[3]{\Pi{((a+b)^2+1)}}\leq 3(\sum{(a+b)^2}+3)=6\sum{a^2}+6\sum{ab}+9\overset{?}{\leq} 8\sum{a^2}+16\sum{ab}\]\[2\sum{a^2}+10\sum{ab}\geq 12\sum{ab}=9\]As desired.$\blacksquare$

Why that $$8(a+b+c)^2 \overset{?}{\geq} 9\sqrt[3]{(1+(a+b)^2)(1+(b+c)^2)(1+(c+a)^2)}$$

Wild wrote: Why that $$8(a+b+c)^2 \overset{?}{\geq} 9\sqrt[3]{(1+(a+b)^2)(1+(b+c)^2)(1+(c+a)^2)}$$ Because taking its third power gives the desired inequality.

Oh yes! l understand thanks

Observe that $(\frac{9}{8})^3=8(ab+bc+ca)^6$ holds. We may rewrite the inequality as $$\left(\frac{a+b+c}{ab+bc+ca}\right)^6\geq 8(1+(a+b)^2)(1+(b+c)^2)(1+(c+a)^2)$$Substitute $x=a+b,y=b+c,z=c+a\Rightarrow ab+bc+ca=\frac{2\sum xy-\sum x^2}{4}$. Plugging in we get $$\left(\frac{2\sum x}{(\sum x)^2-2\sum x^2}\right)^2\geq 2\sqrt[3]{(1+x^2)(1+y^2)(1+z^2)}$$as the new equality. If we apply AM-GM and take the reciprocal of both sides we obtain that if $$\left(\frac{\sum x}{2}-\frac{\sum x^2}{\sum x}\right)^2=\frac{(\sum x)^2}{4}-\sum x^2+\frac{(\sum x^2)^2}{(\sum x)^2}\leq \frac{3}{2\sum x^2+6}...(1)$$is true, the statement is proven. Substitute further and call $\sum x^2=u,(\sum x)^2=v$. Consider the given condition in order to get $$\sum ab=\frac{2\sum xy-\sum x^2}{4}=\frac{3}{4}\Rightarrow v=2u+3$$and $3u\geq v\Rightarrow u\geq 3$. Ultimately, $$(1)\Rightarrow \frac{v}{4}-u+\frac{u^2}{v}\leq \frac{3}{2u+6}\Rightarrow \frac{9u+27}{8u+12}\leq \frac{3}{2}$$which is equavilent to $u\geq 3.$ We're done $\blacksquare$