<s>[INST] 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}$?
 [/INST] Let $a+b=2u$ and $ab=v^2$.
Hence, we need to prove that $3(a^3+b^3)+1-3(1-a-b)\geq\frac{a^2+b^2+1-4(1-a-b)}{a+b}$ or
$$6u(4u^2-3v^2)+1-3(1-2u)\geq\frac{4u^2-2v^2+1-4(1-2u)}{2u},$$
which is a linear inequality of $v^2$, which says that it's enough to prove our inequality for an extremal value of $v^2$, which happens in the following cases.


*

*$v^2\rightarrow0^+$. 


Let $b\rightarrow0^+$. We obtain $3a^3+3a-2\geq\frac{a^2+4a-3}{a}$ or
$$3a^4+2a^2-6a+3\geq0,$$
which is true by AM-GM: $$3a^4+2a^2+3=3a^4+2a^2+4\cdot\frac{3}{4}\geq6\sqrt[6]{3\cdot2\cdot\left(\frac{3}{4}\right)^4}a>6a$$
2. $b=a$, which gives $$6a^3+6a-2\geq\frac{2a^2+8a-3}{2a}$$ or
$$12a^4+10a^2-12a+3\geq0$$ or
$$\left(a^2-\frac{1}{4}\right)^2+\frac{4}{3}\left(a-\frac{3}{8}\right)^2\geq0,$$
which is obvious.
Done!
</s>