Prove $\sum_{\text{cyc}}^{}\sqrt[3] {1+2ac} \le 3\sqrt [3] {3}$. The presented solution of the following problem on Art of Problem Solving using Jensen's inequality is wrong, since the function $f(x):=\sqrt[3]{1+\frac{2t}{x}} $ is convex rather than concave. How would one prove this inequality, correctly?
Let $a, b ,c $ be positive real numbers such that $ a+b+c+abc=4$. Prove that :
$$\sum_{\text{cyc}}^{}\sqrt[3] {1+2ac} \le 3\sqrt [3] {3}.$$
 A: TLDR
A standard computer-assisted (but rigorous) proof is to use Lagrange multipliers method together with interval arithmetic libraries such as IntervalRoots.jl.

We are optimizing within a compact set in $\mathbb R^3$ as shown below

So there exist maximum points, either in the interior, or on the boundaries.
We can use Lagrange method for the interior.
Let
$$
f(a, b, c)=\sqrt[3]{2 a b+1}+\sqrt[3]{2 a c+1}+\sqrt[3]{2 b c+1}-3 \sqrt[3]{3}
$$
and
$$
g(a,b,c,l) = f(a,b,c)+l (a b c+a+b+c-4).
$$
Then we just to find the critical points of $g$, i.e., solve $\nabla g = 0$, i.e., 
\begin{align}
\frac{2 b}{3 (2 a b+1)^{2/3}}+\frac{2 c}{3 (2 a c+1)^{2/3}}+l (b c+1)&=0, \\
\frac{2 a}{3 (2 a b+1)^{2/3}}+l (a c+1)+\frac{2 c}{3 (2 b c+1)^{2/3}}&=0, \\
l (a b+1)+\frac{2 a}{3 (2 a c+1)^{2/3}}+\frac{2 b}{3 (2 b c+1)^{2/3}}&=0, \\
a b c+a+b+c-4&=0.
\end{align}
Look at this for a while and you will see that one solution is
$$
a=b=c=1, l = -\frac{2}{3\ 3^{2/3}}
$$
and this should give us the maximum of $f(1,1,1)=0$.
To rule out other solutions, we can use rigorous numerics libararies like IntervalRoots.jl.
It's not difficult to see that the solution $(a,b,c,l)$ can only be within $[0,4]^3 \times [-55,0]$. The following Julia code finds all such solutions rigorously.
using IntervalArithmetic, IntervalRootFinding, ForwardDiff

f((a, b, c)) = -3*3^(1/3) + (1 + 2*a*b)^(1/3) + (1 + 2*a*c)^(1/3) + (1 + 2*b*c)^(1/3)
g((a, b, c, l))=f((a, b, c))+l*(a + b + c + a*b*c - 4)
∇g = ∇(g)
box = IntervalBox(0..4,3)×(-55..0)
rts = roots(∇g, box, Krawczyk, 1e-5)
println(rts)

And the result is
Root{IntervalBox{4,Float64}}[Root([0.999999, 1.00001] × [0.999999, 1.00001] × [0.999999, 1.00001] × [-0.3205, -0.320499], :unique)]

To see why only chekcing $l \in [-55,0]$ suffices, note that
$$
l = -\frac{2 \left(c (2 a b+1)^{2/3}+b (2 a c+1)^{2/3}\right)}{3 (2 a b+1)^{2/3} (2 a c+1)^{2/3} (b c+1)}
$$
Using $a, b, c \ge 0$ at the bottom and $a, b, c \le 4$ at the top shows that $l > -55$.
This in fact proves (not just verifies) our conjecture that there is only one solution of $\nabla g=0$ (if there is no bug in the library).
However, to be sure that maximum points do not appear on the boundary, we still need to check, for instance $a=0$.
This reduces to find maximum of
$$
\sqrt[3]{2 (4-c) c+1}-3 \sqrt[3]{3}+2,
$$
which is
$$
2-3 \sqrt[3]{3}+3^{2/3} < 0
$$
when $c = 2$.
