Prove $\sum\dfrac{b+c}{a+bc}\ge 2\left[(a+b)(b+c)(c+a)+\frac{3abc}{a+b+c}\right]$ for $ab+bc+ca=1$

Question

Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove$$\dfrac{b+c}{a+bc}+\dfrac{c+a}{b+ca}+\dfrac{a+b}{c+ab}\ge 2\left((a+b)(b+c)(c+a)+\frac{3abc}{a+b+c}\right) $$

I try C-S $$LHS\ge \frac{4(a+b+c)^2}{2+a+b+c-3abc}$$ and we need $$ \frac{2(a+b+c)^2}{2+a+b+c-3abc}\ge(a+b)(b+c)(c+a)+\frac{3abc}{a+b+c} $$ How can I complete the idea ? Thank you.

Answer 1

Proof.

Here is a proof inspired by Nguyen Thai An's idea.

Indeed, by using Cauchy-Schwarz \begin{align*} \frac{b+c}{a+bc}=\frac{b+c}{2a+bc}(a+bc+a)\left(\frac{1}{a+bc}+a\right)-a(b+c)\ge \frac{(b+c)(a+1)^2}{2a+bc}-a(b+c). \end{align*} Also by AM-GM $$2a+bc\le \frac{a}{b+c}+a(b+c)= \frac{a+b+c}{b+c}.$$ Hence,we obtain$$\frac{b+c}{a+bc}\ge \frac{(ab+ac+b+c)^2}{a+b+c}-a(b+c).\tag{*}$$ Let $p=a+b+c; q=ab+bc+ca=1;r=abc.$

Notice that \begin{align*} \sum_{cyc}(ab+ac+b+c)^2&=\sum_{cyc}\left[(ab)^2+(ac)^2+2a^2bc+2a(b^2+c^2)+4abc+b^2+c^2+2bc\right]\\&=2(1-2pr)+2pr+2(p-3r)+12r+2+2(p^2-2)\\&=2p^2-2pr+2p+6r.\tag{**} \end{align*} From $(*)$ and $(**)$ we get$$\dfrac{b+c}{a+bc}+\dfrac{c+a}{b+ca}+\dfrac{a+b}{c+ab}\ge \frac{2p^2-2pr+2p+6r}{p}-2=2(p-r)+\frac{6r}{p}.$$ It means $$\dfrac{b+c}{a+bc}+\dfrac{c+a}{b+ca}+\dfrac{a+b}{c+ab}\ge 2\left[(a+b)(b+c)(c+a)+\frac{3abc}{a+b+c}\right]. $$ The proof is done. Equality holds at $a=b=1;c=0$ and cyclic permutations.

Answer 2

Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$, and $abc=w^3$.

Thus, $v=\frac{1}{\sqrt3}$ and we need to prove that $$\frac{2\cdot9u^2\sqrt3v}{3\sqrt3v^3+9uv^2-3w^3}\geq\frac{9uv^2-w^3}{3\sqrt3v^3}+\frac{w^3}{3uv^2}$$ or $$\frac{6\sqrt3u^2v}{\sqrt3v^3+3uv^2-w^3}\geq\frac{9uv^2-w^3}{3\sqrt3v^3}+\frac{w^3}{3uv^2}$$ or $f(u)\geq0,$ where $$f(u)=27\sqrt3u^3v^4-27u^2v^5+12\sqrt3u^2v^2w^3-6uv^2w^3-3\sqrt3v^4w^3-\sqrt3uw^6+3vw^6.$$ But by Maclaurin($u\geq v\geq w$):
$$f'(u)=81\sqrt3u^2v^4-54uv^5+24\sqrt3uv^2w^3-6v^2w^3-\sqrt3w^6>0,$$ which says that it's enough to prove $f(u)\geq0$ for a minimal value of $u$, which by $uvw$ happens for equality case of two variables.

Now, let $b=a$ and $c=\frac{1-a^2}{2a},$ where $0<a\leq0.$

Thus, we need to prove that: $$(1-a)(9a^9-9a^8+12a^7+42a^6-18a^5+4a^4-16a^3+10a^2-3a+1)\geq0,$$ which is true because if $a=\frac{x}{2}$ we obtain: $$9a^9-9a^8+12a^7+42a^6-18a^5+4a^4-16a^3+10a^2-3a+1=$$ $$=\frac{1}{512}(9x^9-18x^8+48x^7+336x^6-228x^5+128x^4-1024x^3+1280x^2-768x+512)\geq$$ $$\geq\frac{1}{512}(37x^7+336x^6-228x^5+128x^4-1024x^3+896x^2+128)\geq$$ $$\geq\frac{1}{512}(37x^7+208x^6-228x^5+128x^4-768x^3+896x^2)=$$ $$=\frac{x^2}{512}(37x^5+208x^4-228x^3+128x^2-768x+896)\geq$$ $$\geq\frac{x^2}{512}(208x^4-168x^3+128x^2-768x+872)=$$ $$=\frac{x^2}{64}(26x^4-21x^3+16x^2-96x+109)\geq$$ $$\geq\frac{x^2}{64}(37x^2-117x+109)\geq0.$$

Answer 3

pqr method:

Let $p = a + b + c, q = ab + bc + ca = 1, r = abc$. Using $p^2 \ge 3q = 3$, we have $p \ge \sqrt 3$.

We need to prove that $$\frac{2p^2}{2 + p - 3r} \ge p - r + \frac{3r}{p}. \tag{1}$$

If $p \ge 3$, it suffices to prove that $$\frac{2p^2}{2 + p} \ge p$$ which is true.

In the following, assume that $p < 3$.

(1) is written as $$\frac{2p^2}{2 + p - 3r} + \frac{2 + p - 3r}{p} \ge p - r + \frac{2}{p} + 1. $$

By AM-GM, it suffices to prove that $$2\sqrt{\frac{2p^2}{2 + p - 3r} \cdot \frac{2 + p - 3r}{p}} \ge p - r + \frac{2}{p} + 1$$ or $$2\sqrt{2p} \ge p - r + \frac{2}{p} + 1.\tag{2}$$

If $2 \le p < 3$, it suffices to prove that $$2\sqrt{2p} \ge p + \frac{2}{p} + 1 \tag{3}$$ which is true. Indeed, (3) is written as $2\sqrt{2p} - 4 \ge p + \frac{2}{p} - 3$ or $\frac{8(p - 2)}{2\sqrt{2p} + 4} \ge \frac{(p - 1)(p - 2)}{p}$. It suffices to prove that $\frac{8}{2\sqrt{2p} + 4} \ge 1 - \frac{1}{p}$ which is true since $\frac{8}{2\sqrt{2\cdot 3} + 4} \ge 1 - \frac{1}{3}$.

If $\sqrt 3 \le p < 2$, using $r \ge \frac{4pq - p^3}{9} = \frac{4p - p^3}{9}$ (degree three Schur), it suffices to prove that $$2\sqrt{2p} \ge p - \frac{4p - p^3}{9} + \frac{2}{p} + 1 \tag{4}$$ which is true. Indeed, (4) is written as $2\sqrt{2p} - 4 \ge p - \frac{4p - p^3}{9} + \frac{2}{p} - 3$ or $\frac{8(p - 2)}{2\sqrt{2p} + 4} \ge \frac{(p - 2)(p^3 + 2p^2 + 9p - 9)}{9p}$. It suffices to prove that $\frac{8}{2\sqrt{2p} + 4} \le \frac{p^3 + 2p^2 + 9p - 9}{9p} = p^2/9 + 2p/9 + 1 - 1/p$ which is true since $\frac{8}{2\sqrt{2\cdot \sqrt 3} + 4} \le \sqrt 3^2/9 + 2\sqrt 3/9 + 1 - \frac{1}{\sqrt 3}$.

We are done.