Tough multivariable inequality: Minimize $a^2 + b^3 + c^4$ given $a + b^2 + c^3 = \frac{325}{9}$ Let $a,$ $b,$ and $c$ be positive real numbers such that $a + b^2 + c^3 = \frac{325}{9}.$ Find the minimum value of
$a^2 + b^3 + c^4.$

Note that $(a-2)^2+(c^2+2c+3)(c-3)^2+\frac{1}{27}(3b+4)(3b-8)^2\ge 0$ for all $a, b, c\ge 0$, i.e.
$a^2 + b^3 + c^4 + \frac{1093}{27} - 4(a + b^2 + c^3) \ge 0$ for all $a,b,c\ge 0$.
Thus, under the condition $a, b, c\ge 0;\ a+b^2+c^3=\frac{325}{9}$, we have $a^2 + b^3 + c^4 \ge 4\cdot \frac{325}{9} - \frac{1093}{27} = \frac{2807}{27}$.
The minimum is $\frac{2807}{27}$ at $a=2, b= 8/3, c=3$.
