I think the last term is $c\sqrt[3]{a/c}$. Under this, it suffices to prove \[ \sqrt[3]{a^2b}+\sqrt[3]{b^2c}+\sqrt[3]{c^2a}\le ab+bc+ca+\frac23. \]Note that by AM-GM, \[ 3ab + a + \frac13 \ge 3\sqrt[3]{a^2b} \Rightarrow \sqrt[3]{a^2b}+\sqrt[3]{b^2c}+\sqrt[3]{c^2a}\le ab+bc+ca+\frac{a+b+c}{3} + \frac{1}{3} = ab+bc+ca+\frac23. \]

I think you meant $c \sqrt[3]{\frac{a}{c}}$, but yes, thanks for the correction

Let $x=3a, y=3b$ and $z=3c$. Then, $a+b+c=3$ and we need to prove that $\displaystyle \sum 3x\sqrt[3]{\dfrac{y}{x}} \leq \sum xy+6$ By AM-GM, $\displaystyle \sum 3x\sqrt[3]{\dfrac{y}{x}} \leq x(y+\dfrac{1}{x}+1)=\sum xy+\sum x +3=\sum xy+6,$ as desired.

this problem was in the 2nd selection exam of Azerbaijan

Lukaluce wrote: Let $a, b,$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove the following inequality $$a \sqrt[3]{\frac{b}{a}} + b \sqrt[3]{\frac{c}{b}} + c \sqrt[3]{\frac{a}{c}} \le ab + bc + ca + \frac{2}{3}.$$ $$ a \sqrt[3]{\frac{b}{a}} = 3\sqrt[3]{ab \cdot \frac a3 \cdot \frac19}\le ab +\frac a3 + \frac19 $$

A more of a straightforward solution. By Holder inequality, we have $$(ab+bc+ca)(a+b+c)(1+1+1) \ge \left( a \sqrt[3]{\frac{b}{a}} + b \sqrt[3]{\frac{c}{b}} + c \sqrt[3]{\frac{a}{c}} \right)^3$$Let $ab+bc+ca=t$, we have $$\sqrt[3]{3t} \ge a \sqrt[3]{\frac{b}{a}} + b \sqrt[3]{\frac{c}{b}} + c \sqrt[3]{\frac{a}{c}} $$It suffices to prove $t+\frac{2}{3}\ge \sqrt[3]{t}$. Which is equal to, $$(3t+2)^3 \ge 81t$$$$27t^3+54t^2-45t+8\ge 0$$$$(3t-1)^2(3t+8)\ge0$$With the equality holds when $a=b=c=\frac{1}{3}$.

This, is the same, as proving, $f(a^2\cdot b)+f(b^2\cdot c)+f(c^2\cdot a) \le ab+bc+ac+\frac{2}{3},$ where, $f(x)=\sqrt[3]{x}.$ Note, that by AM-GM, $\frac{3ab+a+\frac{1}{3}}{3} \ge \sqrt[3]{a^2\cdot b},$ hence, we have $f(a^2\cdot b)+f(b^2\cdot c)+f(c^2\cdot a) \le ab+bc+ac+\frac{a+b+c+1}{3}=ab+bc+ca+\frac{2}{3},$ as desired.

The original problem is a special case of the following generalization where $$k=3,p=1,\lambda =1$$.

Generalization 1 Let $a,b,c,p,k$ be positive reals such that $a+b+c=1$. Then prove that $$a.\sqrt[k]{\dfrac{b^p} {a^\lambda }}+b.\sqrt[k]{\dfrac{c^p}{b^\lambda }}+c.\sqrt[k]{\dfrac{a^p}{c^\lambda }}\leq \dfrac{3\left(ab+bc+ca\right)}{k}+\dfrac{2\lambda -2p-1}{k}+1$$

Can have a easy solution with Jensen Or just writing $lhs\leq 3*a*b+a+\frac{1}{3}$

Lukaluce wrote: Let $a, b,$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove the following inequality $$a \sqrt[3]{\frac{b}{a}} + b \sqrt[3]{\frac{c}{b}} + c \sqrt[3]{\frac{a}{c}} \le ab + bc + ca + \frac{2}{3}.$$ Proposed by Anastasija Trajanova, Macedonia Generalized Jensen kills it