<s>[INST] Let $a,b,c$ be side lengths of a triangle, $a+b+c=1$. Prove that $P=a^3+b^3+c^3+3abc<\frac{1}{4}$. 
Let $a,b,c$ be side lengths of a triangle such that $a+b+c=1$. Prove that $$a^3+b^3+c^3+3abc<\frac{1}{4}\,.$$

I solved this question. However, I'd like to know if there's a neater solution that doesn't involve the substitutions that I used.
My solution:

$$P=a^3+b^3+c^3-3abc+6abc$$ $$=(\sum a) (\sum a^2 -\sum ab)+6abc$$ $$=\sum a^2 -\sum ab+6abc$$ $$= (\sum a)^2-3\sum ab+6abc$$ $$=1-3\sum ab+6abc.$$
   We have that $abc=4pRr, \sum ab=p^2+r^2+4Rr$ and $p=\frac{1}{2}$ where $p,R,r$ are the semiperimeter, circumradius, and incenter of the triangle, respectively. Plugging these into $P$:
  $$P=1-3p^2-3r^2-12Rr+24pRr$$
$$=\frac{1}{4}-3r^2<\frac{1}{4}.$$ 

 [/INST] A solution without substitutions. 
We need to prove that $$\sum_{cyc}(4a^3+4abc)<(a+b+c)^3$$ or
$$\sum_{cyc}(3a^3-3a^2b-3a^2c+2abc)<0$$ or $$-3\prod_{cyc}(a+b-c)<0,$$ which is obvios.
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