How to prove $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge \frac{3}{2}$ with pqr?

Question

Let $a,b,c\ge 0: a+b+c+3abc=6.$ Prove that $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge \frac{3}{2}$$ I tried to expand $$\frac{2(a+b+c)+ab+bc+ca+3}{abc+ab+bc+ca+a+b+c+1}\ge \frac{3}{2}$$ $$\iff a+b+c+3\ge ab+bc+ca+3abc$$ Replace $a+b+c=6-3abc,$ it suffices to prove $$ab+bc+ca+3abc\le 9$$ I expect to use Schur $abc\ge \dfrac{(a+b+c)(4(ab+bc+ca)-(a+b+c)^2)}{9}$ but I don't know how to transfer it.

I'm looking for a good proof (especially by using Schur ). Also, others idea are welcome.

Thank you

Answer 1

Proof.

WLOG, assuming that $(b-1)(c-1)\ge 0.$ By given hypothesis, we obtain $$abc+a\ge ab+ac \implies 6-a-b-c+3a+b+c\ge (3a+1)(b+c)$$ $$ \implies 0\le b+c\le \frac{2(a+3)}{3a+1}.$$ Now, by using Cauchy-Schwarz \begin{align*} \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}&\ge \frac{1}{a+1}+\frac{4}{b+c+2}\\&=\frac{1}{a+1}+\frac{4}{\dfrac{2(a+3)}{3a+1}+2}\\&=\frac{3}{2}. \end{align*} The proof is done. Equality holds at $a=b=c=1$ and also for $a=b=3;c=0$ and permutations.

Answer 2

By your work we need to prove that: $$ab+ac+bc+6abc\leq9$$ under condition $a+b+c+3abc=6$.

But the condition does not depend on $ab+ac+bc$, which says that it's enough to prove our inequality for the maximal value of $ab+ac+bc$, which by $uvw$(or $pqr$) happens for equality case of two variables.

Let $b=a$ and $c=\frac{6-2a}{1+3a^2}$,where $0\leq a\leq3$, and we need to prove that: $$a^2+\frac{2a(6-2a)}{1+3a^2}+\frac{6a^2(6-2a)}{1+3a^2}\leq9$$ or $$(3-a)(a+1)(a-1)^2\geq0$$ and we are done!

Answer 3

The pqr method:

Let $p = a + b + c, q = ab + bc + ca, r = abc$.

The condition $a + b + c + 3abc = 6$ is written as $$p + 3r = 6. \tag{1}$$

We need to prove that $$p + 3 - q - 3r \ge 0. \tag{2}$$

Using (1), we have $r = 2 - p/3$. (2) is written as $$2p - q - 3 \ge 0. \tag{3}$$

Using $r \ge \frac{4pq - p^3}{9}$ (degree three Schur) and (1), we have $2 - p/3 \ge \frac{4pq - p^3}{9}$ which results in $$q \le \frac{p^3 - 3p + 18}{4p}. \tag{4}$$

From (3) and (4), it suffices to prove that $$2p - \frac{p^3 - 3p + 18}{4p} - 3\ge 0$$ or $$\frac{(6 - p)(p - 3)(p + 1)}{4p}\ge 0$$ which is true using $p \le 6$ and $p \ge 3$ (easy).

We are done.