TITLE: Prove $\sum_{\text{cyc}}^{}\sqrt[3] {1+2ac} \le 3\sqrt [3] {3}$.
QUESTION [6 upvotes]: 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 :
$$\sum_{\text{cyc}}^{}\sqrt[3] {1+2ac} \le 3\sqrt [3] {3}.$$

REPLY [2 votes]: I deleted my previous solution. I give another solution.
WLOG, assume that $c = \min(a,b,c)$. 
Since $x\mapsto \sqrt[3]{x}$ is concave on $(0, \infty)$, we have
\begin{align}
\sqrt[3]{1+2ac} + \sqrt[3]{1+2ba} + \sqrt[3]{1+2cb}
&\le 2\sqrt[3]{\frac{1+2ac + 1+2cb}{2}} + \sqrt[3]{1+2ba}\\
&= 2\sqrt[3]{1+ac + cb} + \sqrt[3]{1+2ba}.
\end{align}
It suffices to prove that 
$$2\sqrt[3]{1+ac + cb} + \sqrt[3]{1+2ba} \le 3\sqrt[3]{3}.$$
We split into two cases:
1) $ba \le 1$: We have $ac + cb \le 2ba\le 2$ and thus
$2\sqrt[3]{1+ac + cb} + \sqrt[3]{1+2ba} \le 3\sqrt[3]{3}$.
2) $ba > 1$: From $a+b+c+abc = 4$, we have
$c = \frac{4-a-b}{ab+1}$. Also, we have $a+b \ge 2\sqrt{ab} > 2$.
Thus, we have $$ac + cb = \frac{(4-a-b)(a+b)}{ab + 1}
= \frac{4 - (a+b - 2)^2}{ab+1} \le \frac{4 - (2\sqrt{ab} - 2)^2}{ab+1}.$$
Thus, it suffices to prove that
$$2\sqrt[3]{1+ \frac{4 - (2\sqrt{ab} - 2)^2}{ab+1}} + \sqrt[3]{1+2ba} \le 3\sqrt[3]{3}.$$
Let $x = \sqrt{ba}$. Then $1 < x \le 2$. It suffices to prove that for $1< x\le 2$,
$$2\sqrt[3]{1+ \frac{4 - (2x - 2)^2}{x^2+1}} + \sqrt[3]{1+2x^2} \le 3\sqrt[3]{3}$$
or
$$2\sqrt[3]{1 + \frac{2(3x-1)(1-x)}{3(x^2+1)}} + \sqrt[3]{\frac{1+2x^2}{3}} \le 3.$$
Note that $\sqrt[3]{1+u} \le 1 + \frac{u}{3}$ for all $u > -1$
since $(1+\frac{u}{3})^3 - (1+u) = \frac{1}{27}u^2(u+9)$.
Thus, it suffices to prove that
$$2\left(1 + \frac{1}{3} \frac{2(3x-1)(1-x)}{3(x^2+1)}\right) +  \sqrt[3]{\frac{1+2x^2}{3}} \le 3 \tag1$$
or
$$\sqrt[3]{\frac{1+2x^2}{3}} \le \frac{21x^2-16x+13}{9(x^2+1)}$$
or
$$\frac{1+2x^2}{3} \le \left(\frac{21x^2-16x+13}{9(x^2+1)}\right)^3$$
or
$$\frac{2(x-1)^2(-243x^6-486x^5+3051x^4-3996x^3+4527x^2-2102x+977)}{729(x^2+1)^3}\ge 0.$$
It is easy to prove that $-243x^6-486x^5+3051x^4-3996x^3+4527x^2-2102x+977 > 0$ for $1 < x\le 2$.
EDIT: Indeed, By letting $x = 1 + v$ for $0 < v \le 1$, we have
\begin{align}
&-243x^6-486x^5+3051x^4-3996x^3+4527x^2-2102x+977\\
=\ & -243v^6-1944v^5-3024v^4-1512v^3+2340v^2+3280v+1728\\
\ge \ & -243v^2-1944v^2-3024v^2-1512v^2+2340v^2+3280v+1728\\
=\ & -4383v^2+3280v+1728\\
\ge \ & -4383v+3280v+1728\\
=\ & -1103v+1728\\
>\ & 0.
\end{align}
We are done.