If $a, b, c, d>0$ such that $a+b+c=1$, prove that $a^3+b^3+c^3+abcd\ge \min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ If $a, b, c, d>0$, such that $a+b+c=1$, prove that: $$a^3+b^3+c^3+abcd\ge \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).$$
I tried solving it as follows:
$$a^3+b^3+c^3=3abc+1-3(ab+bc+ac).$$
From Schur we have that:
$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c).$$
Hence $1+9abc\ge 4(ab+bc+ac)$.
This is as far as I got. I do not know how to introduce the $\min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ to my inequalities. Could you please explain to me how to solve it?

My second solution:
By using Schur's inequality and the identity
$$a^3 + b^3 + c^3 + \frac{15}{4} abc = 
\frac{3}{4}[a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b)] + \frac{1}{4}(a+b+c)^3,$$
we have
$$a^3 + b^3 + c^3 + \frac{15}{4} abc \ge \frac{1}{4}.$$
If $d \ge \frac{15}{4}$, we have
$$a^3 + b^3 + c^3 + abcd \ge a^3 + b^3 + c^3 + \frac{15}{4} abc \ge \frac{1}{4}.$$
If $d < \frac{15}{4}$, we have
\begin{align}
a^3 + b^3 + c^3 + abcd &= a^3 + b^3 + c^3 + \tfrac{15}{4} abc - (\tfrac{15}{4} - d)abc \\
&\ge \frac{1}{4} - (\tfrac{15}{4} - d)(\tfrac{a+b+c}{3})^3\\
&= \frac{1}{4} - (\tfrac{15}{4} - d)(\tfrac{1}{3})^3\\
&= \frac{1}{9} + \frac{d}{27}.
\end{align}
We are done.
