Schur

Since $a,b,c$ are sides of a triangle , so $a,b,c>0$

using concept of weighted means

$\huge{\frac{a^3+b^3+c^3+abc \sqrt{5}}{1+1+1+\sqrt{5}}\geq (a^3 b^3 c^3 (abc)^{\sqrt{5}})^{\frac{1}{3+\sqrt{5}}}}$

$\large{a^3+b^3+c^3+abc \sqrt{5} \geq abc(3+\sqrt{5})}$

now suppose $A \geq B$ and maximum value of $B$ is $C$ so obviously $A\geq C$

$\large{\frac{a+b+c}{3} \geq (abc)^{\frac{1}{3}}}$

$\large{abc \leq 1}$ $\quad as ~a+b+c=3$

thus

$\large{\boxed{a^3+b^3+c^3+abc \sqrt{5} \geq 3+\sqrt{5}}}$