jmc

Let $p,$ $q,$ $r$ be positive constants. Then by AM-GM, $$\begin{array}{rl}{a}^2+p^2& \ge 2pa,\\ b^3+b^3+q^3& \ge 3q{b}^2,\\ c^4+c^4+c^4+r^4& \ge 4r{c}^3.\end{array}$$Hence, $$\begin{array}{rl}{a}^2+p^2& \ge 2pa,\\ 2b^3+q^3& \ge 3q{b}^2,\\ 3c^4+r^4& \ge 4r{c}^3.\end{array}$$Multiplying these inequalities by 6, 3, 2, respectively, we get $$\begin{array}{rl}6a^2+6p^2& \ge 12pa,\\ 6b^3+3q^3& \ge 9q{b}^2,\\ 6c^4+2r^4& \ge 8r{c}^3.\end{array}$$Hence, $$6(a^2+b^3+c^4)+6p^2+3q^3+2r^4\ge 12pa+9q{b}^2+8r{c}^3.(\ast )$$We want to choose constants $p,$ $q,$ and $r$ so that $12pa+9q{b}^2+8r{c}^3$ is a multiple of $a+b^2+c^3.$ In other words, we want $$12p=9q=8r.$$Solving in terms of $p,$ we get $q=\frac{4}{3}p$ and $r=\frac{3}{2}p.$ Also, equality holds in the inequalities above only for $a=p,$ $b=q,$ and $c=r,$ so we want $$p+q^2+r^3=\frac{325}{9}.$$Hence, $$p+\frac{16}{9}{p}^2+\frac{27}{8}{p}^3=\frac{325}{9}.$$This simplifies to $243p^3+128p^2+72p-2600=0,$ which factors as $(p-2)(243p^2+614p+1300)=0.$ The quadratic factor has no positive roots, so $p=2.$ Then $q=\frac{8}{3}$ and $r=3,$ so $(\ast )$ becomes $$6(a^2+b^3+c^4)+\frac{2186}{9}\ge 24(a+b^2+c^3).$$which leads to $$a^2+b^3+c^4\ge \frac{2807}{27}.$$Equality occurs when $a=2,$ $b=\frac{8}{3},$ and $c=3,$ so the minimum value of $a^2+b^3+c^4$ is $\boxed{\frac{2807}{27}}.$