ta có $a^{4}+a^{4}+a^{4}+1\geq 4a^{3}$; TT $3b^{4}+1\geq 4b^{3}$ $\Rightarrow A\geq \frac{4a^{3}+4b^{3}+25c^{3}}{(a+b+c)^{3}}$
có $(4a^{3}+4b^{3}+25c^{3})(\frac{1}{2}+\frac{1}{2}+\frac{1}{5})^{2}\geq (a+b+c)^{3}$ (theo BĐT holder)
$\Rightarrow P\geq \frac{25}{36}$
dấu "=" $\Leftrightarrow a=b=1; c=\frac{2}{5}$