How prove that $ 3(a^3+b^3)+1-3c\ge \frac{a^2+b^2-c^2+1-4c}{a+b}$? Let $ a,b,c>0$ be such that $ a+b+c=1$. How prove that
$ 3(a^3+b^3)+1-3c\ge \frac{a^2+b^2-c^2+1-4c}{a+b}$?

There is no obvious symmetry in this inequality, so let's just try to simplify this with brute force. From $a + b + c = 1$:


*

*$1 - 3c = 3(1 - c) - 2 = 3(a + b) - 2$

*$1 - 4c = 4(1 - c) - 3 = 4(a + b) - 3$

*$c^2 = 1 + a^2 + b^2 -2a - 2b + 2ab$
After replacing these, our goal is now to show:
$3(a^3 + b^3 + a + b) \geq 8 - \frac{4 + 2ab}{a+b} \\
3(a+b)(a^3 + b^3 + a + b) + 2ab + 4 \geq 8(a+b)$
Let's not forget $a + b < 1$. Notice that if $a + b \leq \frac{1}{2}$, we have the result immediately.
Why?
Left hand side is obviously greater than $4 = \frac{8}{2} \geq 8(a+b)$, so from
now on we concentrate on $a + b \in (\frac{1}{2}, 1)$.
We can write $a^3 + b^3 = (a + b)^3 - 3ab(a+b)$, so let's substitute $u = a + b, v = ab$.
Now we have to prove:
$3u^2(u^2 - 3v + 1) + 2v + 4 - 8u \geq 0$
Further instructions: Use calculus. Partial differentiation shows that extrema of this function is located outside of the set we operate in, so any extrema is located on the border of our set.
Show that our function is greater than zero on this border, choose one point in the interior of our set and show that value of the function in it is greater than zero. Now you are done.
I'm curious: where did you find this inequality?
