Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$. Find the minimum of
$$P=\dfrac{1}{1+a}+ \dfrac{1}{1+ab} + \dfrac{1}{3+abc}. $$
My attempt:
I can prove that $\frac{1}{1+t}\geq \frac{5}{9}-\frac{1}{9}t$ and $\frac{1}{3+t}\geq \frac{1}{3}-\frac{1}{9}t$. Now we obtain that
$$P\geq\dfrac{13}{9}-\dfrac{1}{9}\left(a+ab+abc\right)$$
I know the equality holds if $a=2, b =1,c=0$ but now I have no idea to use the assumption of the sum. Help me a hand please. Thank you very much.
Best Answer
For $a=2$, $b=1$ and $c=0$ we obtain: $P=1$.
We'll prove that it's a minimal value.
Indeed, by C-S and AM-GM we obtain: $$P-1=\frac{1}{1+a}+\frac{1^2}{1+ab}+\frac{1^2}{3+abc}-1\geq\frac{1}{1+a}+\frac{4}{4+ab+abc}-1=$$ $$\frac{1}{1+a}-\frac{ab+abc}{4+ab+abc}=\frac{4-a^2b(1+c)}{(1+a)(4+ab+abc)}=$$ $$=\frac{4\left(1-\left(\frac{a}{2}\right)^2b(1+c)\right)}{(1+a)(4+ab+abc)}\geq\frac{4\left(1-\left(\frac{\frac{a}{2}+\frac{a}{2}+b+1+c}{4}\right)^4\right)}{(1+a)(4+ab+abc)}=0.$$