Inequality #2

A little bit of trigonometry: Let a=b=c= tan(60°), not sure if this is valid but 27 is the result for m

Notice that $(a+b)(b+c)(c+a) \geq \dfrac{8(a+b+c)(ab+bc+ca)}{9} \geq \dfrac{8(a+b+c)\sqrt{3abc(a+b+c)}}{9}=\dfrac{8\sqrt{3} a^2b^2c^2}{9}.$ Hence, $(a^3+b^3)(b^3+c^3)(c^3+a^3) \geq ab(a+b)\cdot bc(b+c) \cdot ca(c+a) \geq \dfrac{8\sqrt{3}a^4b^4c^4}{9}.$ Now, using the AM-GM inequality, we have $(ab+1)(bc+1)(ca+1)=\dfrac{(abc+a)(abc+b)(abc+c) }{abc}\leq \dfrac{1}{abc}\left(\dfrac{abc+a+abc+b+abc+c}{3}\right)^3=\dfrac{64a^2b^2c^2}{27}.$ Therefore, $\sum_{cyc}\frac{\sqrt{a^3+b^3}}{ab+1} \geq 3\sqrt[3]{\dfrac{\sqrt{(a^3+b^3)(b^3+c^3)(c^3+a^3)}}{(ab+1)(bc+1)(ca+1)}} \geq 3\sqrt[3]{\dfrac{\sqrt{\dfrac{8\sqrt{3} a^4b^4c^4}{9}}}{\dfrac{64a^4b^4c^4}{27}}}=\dfrac{3\sqrt{2}\sqrt[4]{27}}{4}.$