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 :
$$\sum_{\text{cyc}}^{}\sqrt[3] {1+2ac} \le 3\sqrt [3] {3}.$$
Best Answer
Edit(20230517)
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}. \tag{1} \end{align*}
It suffices to prove that $$2\sqrt[3]{1+ac + cb} + \sqrt[3]{1+2ba} \le 3\sqrt[3]{3}. \tag{2}$$
We split into two cases.
Case 1: $ba < 1$
Using $c = \min(a, b, c)$, we have $ac + cb \le 2ba\le 2$. Thus, (2) is true.
Case 2: $ba \ge 1$
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 $$ac + cb = \frac{(4-a-b)(a+b)}{ab + 1} = \frac{4 - (a+b - 2)^2}{ab+1} \le \frac{4 - (2\sqrt{ab} - 2)^2}{ab+1}. \tag{3}$$
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 $1 \le x \le 2$. It suffices to prove that, for all $1\le x\le 2$, $$2\sqrt[3]{1+ \frac{4 - (2x - 2)^2}{x^2+1}} + \sqrt[3]{1+2x^2} \le 3\sqrt[3]{3}$$ or $$2\sqrt[3]{1 - \frac{2(3x-1)(x-1)}{3(x^2+1)}} + \sqrt[3]{\frac{1+2x^2}{3}} \le 3. \tag{4}$$
Let $$A := 1 - \frac{2(3x-1)(x-1)}{3(x^2+1)}, \quad B := \frac{1+2x^2}{3}.$$ Then $0 < A \le 1$ and $B \ge 1$.
(4) is written as $$\sqrt[3]{B} - 1 \le 2(1 - \sqrt[3]{A})$$ or $$\frac{B - 1}{B^{2/3} + B^{1/3} + 1} \le \frac{2(1 - A)}{A^{2/3} + A^{1/3} + 1}.$$
We have $B^{2/3} + B^{1/3} \ge 2\sqrt{B^{2/3} \cdot B^{1/3}} = 2\sqrt{B} = \frac{4B}{2\sqrt{B}} \ge \frac{4B}{B + 1}$. Also, we have $A^{2/3} + A^{1/3} \le 2A^{1/3} \le 2\cdot \frac{A + 1 + 1}{3} = \frac{2A + 4}{3}$.
Thus, it suffices to prove that $$\frac{B - 1}{\frac{4B}{B+1} + 1} \le \frac{2(1 - A)}{\frac{2A + 4}{3} + 1}$$ or $$\frac{2(x-1)(x+1)(x^2+2)}{15x^2 + 12} \le \frac{12(3x-1)(x-1)}{15x^2 + 16x + 23}$$ or $$\frac{2(x+1)(x^2+2)}{15x^2 + 12} \le \frac{12(3x-1)}{15x^2 + 16x + 23}$$ or (after clearing the denominators) $$-15\,{x}^{4}-46\,{x}^{3}+155\,{x}^{2}-20\,x+118 \ge 0$$ which is true. Indeed, we have $$\mathrm{LHS} \ge -15x^2 \cdot 2^2 - 46x^2 \cdot 2 + 155x^2 - 20 \cdot 2 + 118 = 3x^2 + 78 > 0.$$
We are done.