Can we solve with $AM-GM$

Solution of Zhangyanzong: $$a^3+2\cdot 3^3+5( a^2+3^2)+b^3+2\cdot 4^3+5( b^2+4^2)+c^3+2\cdot 5^3+5( a^2+5^2)\geq 57a+88b+125c$$$$a^3 + b^3 + c^3 + 5a^2 + 5b^2 + 5c^2\geq 466$$Equality holds when $a=3,b=4,c=5.$

Let $f(x)=x^3+5x^2$, so that we seek to minimize $f(a)+f(b)+f(c)$ over positive reals $a$, $b$, $c$ obeying $57a+88b+125c\ge1148$. Check that $f'(x)=3x^2+10x$ and $f''(x)=6x+10$, so $f$ is convex over $\mathbb R^+$. From this, it is clear that $x^3+5x^2\ge57x-99$, which is the tangent to $f$ at $(3,72)$. $x^3+5x^2\ge88x-208$, which is the tangent to $f$ at $(4,144)$. $x^3+5x^2\ge125x-375$, which is the tangent to $f$ at $(5,250)$. Summing these, $f(a)+f(b)+f(c)\ge1148-99-208-375=466$, achieved by $(a,b,c)=(3,4,5)$. Remark. The motivation for such a solution is the intuition that we should be able to find a single-variable inequality of the form $f(x)=x^3+5x^2\ge px+q$, and cyclically sum. Indeed, we want the inequalities $f(a)\ge 57a+q_1$, $f(b)\ge 88b+q_2$, and $f(c)\ge 125c+q_3$. Since $f$ is convex, we want these to be tangents to $f$. We find that the point on the graph of $f$ with slope $57$ has $x$-coordinate $3$ (say, by solving $0=3x^2-10x-57=(3x+19)(x-3)\implies x=3$). Miraculously the other two also yield integer $x$-coordinates, and the result is clear from there.

Let $f(x)=x^3+5x^2$, then we find out that $f'(3)=57$, $f'(4)=88$, $f'(5)=125$. We have the relations $f(x)-f(3)\ge f'(3)(x-3)$, et cetera, which gives $$f(a)+f(b)+f(c) \ge f(3)+f(4)+f(5)+(57a+88b+125c)-3\cdot 57-4\cdot 88-5\cdot 125 \ge 466.$$

Idio-logy wrote: We have the relations $f(x)-f(3)\ge f'(3)(x-3)$, et cetera, Sorry, I don't see why the values of the derivative imply this fact. Could you please elaborate?

Just factorize the expression; you can see $f'(3)(x-3)+f(3)$ as the tangent line to $f(x)$ at $x=3$, and $f(x)\ge f'(3)(x-3)+f(3)$ manages to hold for all $x\ge 0$.

We have the following inequalities over the positive reals: \begin{align*} (a-3)^2(a+11) \geq 0 &\Rightarrow a^3+5a^2 \geq 57a-99 \\ (b-4)^2(b+13) \geq 0 &\Rightarrow b^3+5b^2 \geq 88b-208 \\ (c-5)^2(c+15) \geq 0 &\Rightarrow c^3+5c^2 \geq 125c-375 \\ \end{align*}. Summing these gives $a^3+b^3+c^3+5(a^2+b^2+c^2) \geq \boxed{466}$, with equality when $a=3,b=4, c=5$.

Heh, had a partial derivatives lesson in multivariable calculus class a few days before doing this question :p. First of all, since $57a+88b+125c \ge 1148$ and we need the minimum value of $\sum_{\text{cyc}} (a^3+5a^2)$, then that gets us that for the minimum value, we can assume $57a+88b+125c = 1148$. Now, due to that, we see that letting $57a+88b+125c = F(a, b, c)$ for some constant $d$, applying partial derivatives gets that $F_a = 57$, $F_b = 88$, and $F_c = 125$. Now, this means that $\frac{G_a}{G_b} = \frac{F_a}{F_b} = \frac{57}{88}$ and similar for $\frac{G_b}{G_c}=\frac{F_b}{F_c}=\frac{88}{125}$ and $\frac{G_c}{G_a}=\frac{F_c}{F_a}=\frac{125}{57}$. We notice $(a, b, c) = (3, 4, 5)$ satisfies this condition and for this pair, $57a+88b+125c = 1148$. All other points with all three coordinates positive result in at least one partial derivative being nonzero meaning they cannot be an extreme point, and since taking a random point on the plane $57a+88b+125c = 1148$ gets us a value greater than $\sum_{\text{cyc}} (a^3+5a^2)$ when $a=3$, $b=4$ and $c=5$, then our answer for the minimum value is just $3^2 \cdot 8 + 4^2 \cdot 9 + 5^2 \cdot 10 = \boxed{466}$.

EQUALITY!!!!!!!!!!!!!!!!!!!!!!!!! Equality holds at $(3, 4, 5)$ which gives us a $466$ minimum. From here, observe the magical inequality $a^3 + 5a^2 + 57a - 243 \geq 0$, which is because this is just $(a - 3)^2 (a + 11)$. Similarly, we get $(b - 4)^2(b + 13) \geq 0$ and $(c - 5)^2(c + 15) \geq 0$ which all have the correct quadratic and linear coefficients. Cyclically sum to finish.

kevinmathz wrote: Heh, had a partial derivatives lesson in multivariable calculus class a few days before doing this question :p. First of all, since $57a+88b+125c \ge 1148$ and we need the minimum value of $\sum_{\text{cyc}} (a^3+5a^2)$, then that gets us that for the minimum value, we can assume $57a+88b+125c = 1148$. Now, due to that, we see that letting $57a+88b+125c = F(a, b, c)$ for some constant $d$, applying partial derivatives gets that $F_a = 57$, $F_b = 88$, and $F_c = 125$. Now, this means that $\frac{G_a}{G_b} = \frac{F_a}{F_b} = \frac{57}{88}$ and similar for $\frac{G_b}{G_c}=\frac{F_b}{F_c}=\frac{88}{125}$ and $\frac{G_c}{G_a}=\frac{F_c}{F_a}=\frac{125}{57}$. We notice $(a, b, c) = (3, 4, 5)$ satisfies this condition and for this pair, $57a+88b+125c = 1148$. All other points with all three coordinates positive result in at least one partial derivative being nonzero meaning they cannot be an extreme point, and since taking a random point on the plane $57a+88b+125c = 1148$ gets us a value greater than $\sum_{\text{cyc}} (a^3+5a^2)$ when $a=3$, $b=4$ and $c=5$, then our answer for the minimum value is just $3^2 \cdot 8 + 4^2 \cdot 9 + 5^2 \cdot 10 = \boxed{466}$. $G$ stand for ...?

By the tangent line trick, we derive the inequalities \[a^3 + 5a^2 \ge 72 + 57(a-3) \Longleftrightarrow (a-3)^2(a+11) \ge 0\] \[b^3 + 5b^2 \ge 144 + 88(b-4) \Longleftrightarrow (b-4)^4(b+13)\] and \[c^3 + 5c^2 \ge 250 + 125(c-5) \Longleftrightarrow (c-5)^2(c+15) \ge 0.\] Then summing these yields \begin{align*} a^3 + b^3 + c^3 + 5(a^2 + b^2 + c^2) & \ge 466 + 57(a-3) + 88(b-4) + 125(c-5)\\ & = 466 + 57a + 88b + 125c - 1148\\ & \ge 466 \end{align*} yielding an answer of $\boxed{466}$. Equality occurs at $(a, b, c) = (3, 4, 5)$, so this value is indeed attainable. $\square$

The minimum is $466$ obtained at $(a,b,c)=(3,4,5)$, which we prove by adding the inequalities:

The minimum is $466$ at $(3, 4, 5)$. Indeed we can just write the three expressions \begin{align*} a^3+5a^2 &\geq 57a+99 \\ b^3+5b^2 &\geq 88b+208 \\ c^3+5c^3 &\geq 125c+375. \end{align*}

This problem was written by Satan (or any other corresponding deity). WLOG we can assume that the equality $57a + 88b + 125c = 1148$ holds. Define $f(x) = x^2 + 5x^2$. Then we want to minimize $f(a) + f(b) + f(c)$ under the equality constraint of $f'(3)a + f'(4)b + f'(5)c = 1148$. Define $g(a, b, c) = f'(3)a + f'(4)b + f'(5)c - 1148$ which has nonzero gradient, and $h(a, b, c) = f(a) + f(b) + f(c)$. Since this is a compact set, minimality must hold by LM when \[ \nabla_h = \lambda \nabla_g \]or \[ (f'(x), f'(y), f'(z)) = (\lambda f'(3), \lambda f'(4), \lambda f'(5)) \]The equality case can only hold when $x = 3, y = 4, z = 5$ however as $f'(x)$ is monotonic, giving $466$ as the answer.

Let $f(x) = x^3+5x^2$. From the Tangent Line Trick, we get \[f(x) \ge f(k) + (x-k) \cdot f'(k) = (3k^2+10k)x - (2k^3+5k^2).\] Computing for $k=3,4,5$, we get our bound \[f(a)+f(b)+f(c) \ge 57a+88b+125c-(99+208+375) = \boxed{466},\] which can be obtained with $(a,b,c)=(3,4,5)$. $\blacksquare$

Note that \[(a-3)^2(a+11)+(b-4)^2(b+13)+(c-5)^2(c+15)\ge 0,\]and expanding, \[a^3+b^3+c^3+5(a^2+b^2+c^2)-(57a+88b+125c)+99+208+375\ge 0.\]So the desired expression is at least $1148-99-208-375=466$, achieved by $(a,b,c)=(3,4,5)$. $\blacksquare$