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?

WLOG, assume $c = \min(a, b, c)$.
Let $P(x) = a^3 + b^3 + c^3 + abc x$ and $Q(x) = 2(\frac{a+b}{2})^3 + c^3 + (\frac{a+b}{2})^2 c x$ where $x \ge 0$.
We have
\begin{align}
P(x) - Q(x) &= a^3 + b^3 - 2(\tfrac{a+b}{2})^3
- ((\tfrac{a+b}{2})^2 - ab)c x \\
&= \frac{3}{4}(a+b)(a-b)^2 - \frac{1}{4}(a-b)^2 cx\\
&= \frac{1}{4}(a-b)^2(3a + 3b - cx)\\
&\ge \frac{1}{4}(a-b)^2(3a + 3b - \tfrac{a+b}{2} x)\\
&= \frac{1}{8}(a-b)^2(a+b)(6 - x). \tag{1}
\end{align}
Also, since $a + b + c = 1$, we have $Q(x) = \frac{1}{4}(x+3)c^3 + \frac{1}{4}(-2x+3)c^2 + \frac{1}{4}(x-3)c + \frac{1}{4}$.
Now, we split into two cases:
If $d > \frac{15}{4}$, by (1), we have
\begin{align}
a^3 + b^3 + c^3 + abcd &\ge a^3 + b^3 + c^3 + \frac{15}{4} abc\\
&= P(\tfrac{15}{4}) \\
&\ge Q(\tfrac{15}{4}) \\
&= \frac{1}{4}(\tfrac{15}{4} +3)c^3 + \frac{1}{4}(-2\cdot \tfrac{15}{4} + 3)c^2 + \frac{1}{4}(\tfrac{15}{4}-3)c + \frac{1}{4}\\
&= \frac{3}{16}c(3c-1)^2 + \frac{1}{4}\\
&\ge \frac{1}{4}\\
&= \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).
\end{align}
If $d\le \frac{15}{4}$, by (1), we have
\begin{align}
a^3 + b^3 + c^3 + abcd &= P(d) \\
&\ge Q(d)\\
&= \frac{1}{4}(d+3)c^3 + \frac{1}{4}(-2d+3)c^2 + \frac{1}{4}(d-3)c + \frac{1}{4}\\
&= \frac{1}{9} + \frac{1}{27}d + \frac{1}{108}(3c-1)^2(3cd + 9c - 4d + 15)\\
&\ge \frac{1}{9} + \frac{1}{27}d\\
&= \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).
\end{align}
We are done.
