Prove $(a^3+b^3+c^3)^2 \ge 3(a^4 + b^4 + c^4)$ for $abc = 1$ ($a, b, c > 0$)

Question

Let $a,b,c$ be positive real numbers such that $abc=1$. Does it then hold that $$(a^3+b^3+c^3)^2 \ge 3(a^4 + b^4 + c^4)$$

A lazy and non-rigorous computation suggests that, after substituting $c = \frac {1}{ab}$, the resulting function in $a,b$ is convex, and hence the local minimum at $a = b = 1$ is the global minimum and the inequality is true.

This problem came out of an amusing game I have been playing lately, in which I ask ChatGPT to make up Putnam problems which I then attempt (they are often hilariously false). The above question was one such problem which my basic knowledge of olympiad inequalities was unable to make much headway in either direction on (my ideas to use mean inequalities or symmetric polynomial inequalities failed, though I did not try Lagrange multipliers or similar methods). If the AI stole it from somewhere I would like to know.

Answer 1

I came up with a proof in Jan 10, 2023.

We have \begin{align*} &(a^3 + b^3 + c^3)^2 - 3(a^4 + b^4 + c^4)\\ ={}& a^6 + b^6 + c^6 + 2a^3b^3 + 2b^3c^3 + 2c^3a^3 - 3(a^4 + b^4 + c^4)\\ \ge{}& a^6 + b^6 + c^6 + (3a^2b^2 - 1) + (3b^2c^2 - 1) + (3c^2a^2 - 1) - 3(a^4 + b^4 + c^4) \tag{1}\\ ={}& a^6 + b^6 + c^6 + 3a^2b^2 + 3b^2c^2 + 3c^2a^2 - 3a^4 - 3b^4 - 3c^4 - 3\\ ={}& a^6 + b^6 + c^6 + 3a^2b^2 + 3b^2c^2 + 3c^2a^2 - 3a^4 - 3b^4 - 3c^4 - 3a^2b^2c^2\\ ={}& (a^2 + b^2 + c^2 - 3)(a^4 + b^4 + c^4 -a^2b^2 - b^2c^2 - c^2a^2)\\ \ge{}& 0 \end{align*} where we use $2a^3b^3 - (3a^2b^2 - 1) = (2ab + 1)(ab - 1)^2 \ge 0$ in (1), and $a^2 + b^2 + c^2 \ge 3$ by AM-GM, and $a^4 + b^4 + c^4 -a^2b^2 - b^2c^2 - c^2a^2 \ge 0$ (easy).

We are done.

Answer 2

I doubt if this inequality is true. As per Wolfram Alpha, it does not have a global minima.

Take $a = x/y, b = y/z$ and $c = z/x$. Then we want to show that

$$ \Big(\Big(\frac{x}{y}\Big)^3 + \Big(\frac{y}{z}\Big)^3 + \Big(\frac{z}{x}\Big)^3 \Big)^2 - 3\Big(\Big(\frac{x}{y}\Big)^4 + \Big(\frac{y}{z}\Big)^4 + \Big(\frac{z}{x}\Big)^4 \Big) \ge 0 $$

But Wolfram Alpha says the expression on the left has no global minima. Further more even if we relax the condition that $a,b,c$ must be positive, Wolfram Alpha still says that there is no global minima.