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?

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 $$f(a,b,c)=2a^3+(1-2a)^3+a^2(1-2a)\left(\frac3a-3\right)=3a^2-3a+1\ge\frac14.$$
