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 $P$:
  $$P=1-3p^2-3r^2-12Rr+24pRr$$
$$=\frac{1}{4}-3r^2<\frac{1}{4}.$$ 

 A: 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.
A: Because it is a somewhat rare occurrence that the inequality could be derived easily/naively from the equality case, I'm posting this solution which is essentially the same as Michael's written backwards.    
Observe that when $ a = b = 0.25, c = 0.5$, we get equality.
This suggests that we need to look at the triangle inequality $ a + b - c > 0 $, and put it into play.
And of course, cyclic versions of it.   
Naively, we look at $ \prod ( a + b - c ) > 0$.   
Expanding this out, we get 
$$ a^3 + b^3 + c^3 + 2abc < a^2b + b^2 c + c^2 a + a^2 c + b^2 a + c^2 b $$
This looks very promising, given that we've bounded the high power terms.   
Of course, the next step is to use $ a + b + c = 1$. Naturally, we cube this to get   
$$ a^3 + b^3 + c^3 + 3a^2b + 3a^2 c + 3b^2 c + 3b^2 a + 3c^2 a + 3c^2 b + 6 abc = 1.$$ 
This only uses terms that appear on both sides, so that's again very promising. Put them together to get 
$$ a^3 + b^3 + c^3 + 2abc < a^2b + b^2 c + c^2 a + a^2 c + b^2 a + c^2 b  = \frac{ 1 - (a^3 + b^3 + c^3  + 6abc)}{3} $$
Hence, we obtain  
$$ 4a^3 + 4b^3 + 4c^3 + 12 abc < 1$$

We might have considered $ \sum \prod(a+b+c)(a+b-c)(a-b+c) > 0$, which expands to the weaker inequality 
$$ a^3 + b^3 + c^3 - 6abc < a^2b+ b^2 c + c^2a + a^2c + b^2a + c^2 b. $$
Thinking about why it's so much weaker, this is because only 1 of the triangle inequalities can become an equality for fixed values, so this inequality can never become an equality (in the limiting case).   
So, we need all 3 triangle inequalities to be in play at the same time, which is why $ \prod ( a + b - c)$ is so natural to consider. 
