Tough multivariable inequality: Minimize $a^2 + b^3 + c^4$ given $a + b^2 + c^3 = \frac{325}{9}$

Question

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

Answer 1

Note that for the surface $a^2 + b^3 + c^4=k$ to have the minimum value $k$, it is tangential to the surface $a + b^2 + c^3 = \frac{325}{9}$. Calculate their normal vectors $(1,2b,3c^2)$ and $(2a,3b^2,4c^3)$, and match them to get

$$\frac{2a}1=\frac{3b}2=\frac{4c}3$$

Then, plug $b= \frac43a$ and $c=\frac32a$ into $a + b^2 + c^3 = \frac{325}{9}$ to obtain the tangential point at

$$a=2,\>\>\>b = \frac83,\>\>\>c=3$$

As a result, the minimum value is

$$a^2 + b^3 + c^4=2^2+\left(\frac83\right)^3+3^4=\frac{2807}{27}$$

Answer 2

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 $a=2, b= 8/3, c=3$.

Answer 3

Our problem reduces to $$\min (x^2 + y^{\frac{3}{2}}+ z^{\frac{4}{3}})$$ when $x$, $y$, $z \ge 0$, $x+y+z = \frac{325}{9}$.

We sketch an approach to the general problem $$\min \sum_{i=1}^n x_i^{a_i} $$ where $x_i\ge 0$ and $\sum x_i = s$. ($a_i>1$, $s>0$ fixed).

For $a>1$ and $\lambda>0$ consider the function $$f_{a, \lambda}(x)= x^{a} - \lambda x$$ on $[0, \infty)$. We have $$f_{a,\lambda}^{''}(x)= a(a-1)x^{a-2}$$ so $f_{a,\lambda}$ is strictly convex on $(0,\infty)$. Therefore, $f_{a,\lambda}$ has a unique minimum $m_{a, \lambda}$ on $[0, \infty)$ at the point where the derivative $=0$, that is, at $$x_{a,\lambda} = \left(\frac{\lambda}{a}\right)^{\frac{1}{a-1}}$$

Therefore, we have the inequality $$ x^a - \lambda x \ge m_{a, \lambda} $$

Summing up for $i=1, \ldots, n$ we get $$\sum x_i^{a_i} \ge \lambda \sum x_i + \sum_i m_{a_i, \lambda} $$

Now, every $\lambda\ge0$ the above gives a lower bound for $\sum x_i^{a_i}$ in terms of $\sum x_i=s$. To get an optimal bound, we need that the equality be achievable. So we need to have $$\sum x_{a_i, \lambda} = s$$ that is $$\sum_i \left(\frac{\lambda}{a_i}\right)^{\frac{1}{a_i-1}}=s$$ This equation in $\lambda$ has a unique solution $\bar \lambda =\lambda_{(a_i), s}$.

So we get $$\min \sum x_i^{a_i}= \bar \lambda \cdot s + \sum m_{a_i, \bar\lambda}$$

Generalization: minimize a sum of convex functions $\sum f_i(x_i)$ for a given $\sum x_i$.

Answer 4

Have you tried considering the Lagrangean $$\mathcal L(a, b, c) = a^2 + b^3 + c^4 + \lambda\left(a + b^2 + c^3 - \frac{325}9\right)?$$