ii. If $a, b, c$ are positive real numbers satisfying $a+b+c=1$, find $f(a, b, c)=\frac{a^{4}}{a^{3}+b^{2}+c^{2}}+\frac{b^{4}}{b^{3}+a^{2}+c^{2}}+\frac{c^{4}}{c^{3}+b^{2}+a^{2}}$
Let $a=b=c=\frac{1}{3}$, then $f(a, b, c)=\frac{1}{7}$. We need to prove: $f(a, b, c) \geq \frac{1}{7}$;
Let $x=a b+b c+c a, y=a b c$, then $a^{2}+b^{2}+c^{2}=1-2 x$.
$$
\text { Since } \begin{aligned}
2\left(a^{3}+b^{3}+c^{3}\right) & =3\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)+6 a b c-(a+b+c)^{3} \\
& =3(1-2 x)+6 y-1=2(1-3 x+3 y)
\end{aligned}
$$

where $x \leq \frac{1}{3}, 3 y \leq x^{2}$. By the Cauchy-Schwarz inequality, we have:
$$
\frac{a^{4}}{a^{3}+b^{2}+c^{2}}+\frac{b^{4}}{b^{3}+a^{2}+c^{2}}+\frac{c^{4}}{c^{3}+b^{2}+a^{2}} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a^{3}+b^{3}+c^{3}+2\left(a^{2}+b^{2}+c^{2}\right)}=\frac{(1-2 x)^{2}}{3-7 x+3 y},
$$

Therefore, it suffices to prove: $7(1-2 x)^{2} \geq 3-7 x+3 y$, which is equivalent to proving: $(3 x-1)(9 x-4) \geq 0$ (*); Since $x \leq \frac{1}{3}$, (*) is obviously true.