If $a, b, c, d>0$ such that $a+b+c=1$, prove that $a^3+b^3+c^3+abcd\ge \min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$

This is an easy application of Lagrange as the first equality of $$3a^2+bcd=3b^2+acd=3c^2+abd=\lambda$$ gives $3(a+b)(a-b)=(a-b)cd$ so either $a=b$ or $3(a+b)=cd$. Thus there are two possibilities:

• $a=b=c$ so that $f(a,b,c)=1/9+d/27$;

• $a=b$ and $3(b+c)=ad$ so that $d=3/a-3$ since $c=1-2a$ giving $$f(a,b,c)=2a^3+(1-2a)^3+a^2(1-2a)\left(\frac3a-3\right)=3a^2-3a+1\ge\frac14.$$

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.