Funny problem (Quite easy after you have seen weird inequality conditions and try to get weird identity like TSTST 2012 #06). Here's a solution with motivation. Let's deal with $a + b + c = ab + bc + ca$ and try to find a nice bound. $\textbf{Claim.}$ $a + b + c \ge 3$. To prove this, notice that \[ (a+b+c)(ab + bc + ca) = (a+b+c)^2 \ge 3(ab + bc + ca) \] The major problem here is dealing with $\sqrt[3]{\frac{a^3 + 1}{2}}$. This motivates us to consider AM-GM in the form like \[ \sqrt[3]{ \left( \frac{a + 1}{2} \right)(a^2 - a + 1)} \]by considering the equality of AM-GM. Now, notice the fabulous identity \[ a^2 - a + 1 = a^2 + ab + bc + ca + 1 - 2a - b - c = (a+b)(a + c) + 1 - 2a - b - c = (a + b)(a+c) + 1 - ((a+b) + (a+c)) = (a+b - 1)(a + c - 1)\]Therefore, we have \begin{align*} \sqrt[3]{\frac{a^3 + 1}{2}} &= \sqrt[3]{ \left( \frac{a + 1}{2} \right) (a^2 - a + 1)} \\ &= \sqrt[3]{\left( \frac{a + 1}{2} \right)(a + b - 1)(a + c - 1)} \\ &= \frac{1}{3} \left( \frac{a + 1}{2} + (a + b - 1) + (a + c - 1) \right) \\ &= \frac{1}{3} \left( \frac{5}{2} a + b + c - \frac{3}{2} \right) \end{align*}Now, sum cyclically and we get \begin{align*} 3 + \sum_{cyc} \sqrt[3]{\frac{a^3 + 1}{2}} &\le 3 + \sum_{cyc} \frac{1}{3} \left( \frac{5}{2} a + b + c - \frac{3}{2} \right) \\ &= 3 + \frac{3}{2} (a + b + c) - \frac{3}{2} \\ &= \frac{3}{2} (a + b + c + 1) \\ &\le 2( a + b + c) \end{align*}$\textbf{Remark.}$ We need to prove $a + b, a + c, b + c > 1$ for any $a,b,c$ satisfying the statement of the problem. Suppose otherwise, and WLOG $a \le b \le c$. We then have $a + b \le 1$. Now, \[ c(a+b) + ab = (a+b) + c \rightarrow c(1 - (a+b)) = ab - (a + b) = (1 - a)(1 - b) - 1 \le 0 \]since both $1 - a$ and $1 -b$ are less than $1$, and positive. However, this forces $c \le 0$ or $a + b = 1$. But the latter one force $ab = a + b = 1$, which doesn't has a solution since $ab \le \frac{1}{4}$.

Nice solution @above . Here is a more direct approach by using Jensen. Note that the function $x \mapsto x^{\frac 13}$ is concave so we have $:=$ $$ \sum_{cyc} \sqrt[3] {\frac {a^3+1}{2}} \leq 3\sqrt[3] {\frac {a^3+b^3+c^3+3}{6}}$$ We now wish to prove $:=$ $$\sqrt[3] {\frac {a^3+b^3+c^3+3}{6}} \leq \frac {2(a+b+c)-3}{3}$$ Write $s=a+b+c$ and $p=abc$. Note that $(a+b+c)^2\geq 3(ab+bc+ca) \implies s\geq 3$. Writing everything in $s,p$ , we want : $$ \frac {s^3-3s^2+3p+3}{6} \leq \frac {(2s-3)^3}{27}$$After expanding, this is equivalent to : $$7s^3-45s^2+108s-27p-81 \geq 0$$Using $s^3 \geq 27p$ , this is equivalent to proving: $$0 \leq 6s^3-45s^2+108s-81=3(s-3)^2(2s-3)$$which is evident as $s \geq 3$. We are done. $\blacksquare$

By Mildorf's Lemma, $\sqrt[3]{\dfrac{a^3+1}{2}} \leqslant \dfrac{a^2+1}{a+1}$, and the cyclic ones. So, $\displaystyle LHS \leqslant \sum \dfrac{a^2+1}{a+1}+3$. But, $\sum \dfrac{a^2+1}{a+1}=\sum (a-1)+\sum \dfrac{2}{a+1}$, hence it suffices to show $\sum \dfrac{2}{a+1} \leqslant \sum a$. Let $\sum a=\sum ab=p$ and $abc=r$. Then expanding the previous one we have to prove that $r \geqslant \dfrac{5p+6-2p^2}{p}$. If $5p+6 \leqslant p^2$, we are trivially done. On the other hand, if $5p+6 >p^2$, then $p<4$. By Schur, $p^3+9r \geqslant 4p^2$, so $r \geqslant \dfrac{4p^2-p^3}{9}$. We can easily now verify that $\dfrac{4p^2-p^3}{9} \geqslant \dfrac{5p+6-2p^2}{p}$, for all $p \in [3,4)$ (recall that $p=\sum ab \leqslant p^2/3$, so $p \geqslant 3$). Equality holds when $a=b=c=1$.

Can you explain about mildorf's theorem?

Trick from Poland 2007 and I believe it had been known to some even earlier.

Hint

Multiply the left-hand side by $ab+bc+ca$ and the right-hand side by $a+b+c$. Subtracting $3(ab+bc+ca)$ from both sides, we want to prove the equivalent inequality: \[(ab+bc+ca) \sum\limits_{cyc} \sqrt[3]{\frac{a^3+1}{2}} \leq 2\sum\limits_{cyc} a^2 + \sum\limits_{cyc} ab\]Miraculously, we have: \begin{align*} (ab+bc+ca)\sum\limits_{cyc}\sqrt[3]{\frac{a^3+1}{2}} &= \sum\limits_{cyc}\sqrt[3]{\frac{1}{2}(a^3(a+b+c)^3+(ab+bc+ca)^3)} \\ &=\sum\limits_{cyc}\sqrt[3]{\frac{1}{2}(a^2+2ab+2ac+bc)(a^2+ab+b^2)(a^2+ac+c^2)} \\ &\leq \sum\limits_{cyc} \frac{1}{3}\left(\frac{1}{2}(a^2+2ab+2ac+bc)+ (a^2+ab+b^2) + (a^2+ac+c^2)\right) \\ &= \frac{3}{2}\sum\limits_{cyc} a^2 + \frac{3}{2}\sum\limits_{cyc} ab \\ &\leq 2\sum\limits_{cyc} a^2 + \sum\limits_{cyc} ab \end{align*}

Consider $a^2-a+1=(a+b-1)(a+c-1)$ Then $3+\sum \sqrt[3]{\frac{a^3+1}{2}}= 3+\sum \sqrt[3]{\frac{a+1}{2}(a+b-1)(a+c-1)}\le 3+\sum \frac{\frac{5a}{2}+b+c-\frac{3}{2}}{3}$ sum cyc get result