If $ab+bc+ca=3$, then prove that $\sum\sqrt{\frac{b+c}{bc+1}}\ge\frac{a+b+c+3}{\sqrt{a+b+c+abc}}$

Question

Problem Let $a,b,c\ge 0: ab+bc+ca=3$. Prove that$$\sqrt{\frac{b+c}{bc+1}}+\sqrt{\frac{c+a}{ca+1}}+\sqrt{\frac{a+b}{ab+1}}\ge\frac{a+b+c+3}{\sqrt{a+b+c+abc}}.$$ I saw the problem on AOPS.

I tried to use AM-GM which leads to wrong inequality$$3.\sqrt[3]{\sqrt{\frac{b+c}{bc+1}}.\sqrt{\frac{c+a}{ca+1}}.\sqrt{\frac{a+b}{ab+1}}}\ge\frac{a+b+c+3}{\sqrt{a+b+c+abc}}.$$ How can I prove it?

Answer 1

Another way.

By C-S $$\sum_{cyc}\sqrt{\frac{a+b}{ab+1}}=\sum_{cyc}\sqrt{\frac{(a+b)(a+b+c+abc)}{(ab+1)(a+b+c+abc)}}=\sum_{cyc}\sqrt{\frac{\frac{(a+b)^2}{ab+1}+c(a+b)}{a+b+c+abc}}=$$ $$=\sum_{cyc}\sqrt{\tfrac{\frac{(a+b)^2}{ab+1}+\frac{c^2(a+b)^2}{c(a+b)}}{a+b+c+abc}}\geq\sum_{cyc}\sqrt{\tfrac{\frac{(a+b+c(a+b))^2}{ab+1+c(a+b)}}{a+b+c+abc}}=\sum_{cyc}\tfrac{a+b+c(a+b)}{2\sqrt{a+b+c+abc}}=\tfrac{a+b+c+3}{\sqrt{a+b+c+abc}}.$$

Answer 2

By Holder $$\sum_{cyc}\sqrt{\frac{a+b}{ab+1}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{a+b}{ab+1}}\right)^2\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+4c)^3}{\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+4c)^3}}\geq$$ $$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(a+b)(a+b+4c)\right)^3}{\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+4c)^3}}=\sqrt{\frac{8\left(\sum\limits_{cyc}(a^2+5ab)\right)^3}{\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+4c)^3}}$$ and it's enough to prove that: $$8\left(\sum\limits_{cyc}(a^2+5ab)\right)^3(a+b+c+abc)\geq(a+b+c+3)^2\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+4c)^3.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$,where $v>0$(we obtain $v=1$), $abc=w^3$

and for any polynomial $p(x)$ let $C(p)$ it's a coefficient before greatest power of $x$.

We need to prove that $$8(9u^2+9v^2)^3(3uv^2+w^3)\geq(3u+3v)^2\sum_{cyc}(a+b)^2(ab+v^2)(3u+3c)^3$$ or $f(w^3)\geq0,$ where $$f(w^3)=24(u^2+v^2)^3(3uv^2+w^3)-(u+v)^2\sum_{cyc}(a+b)^2(ab+v^2)(u+c)^3.$$ But $$C(f)=-(u+v)^2C\left(\sum_{cyc}(a+b)^2(ab+v^2)(u+c)^3\right)=-(u+v)^2C\left(\sum_{cyc}(3u-c)^2ab(u+c)^3\right)=$$ $$=-(u+v)^2C\left(\sum_{cyc}(9u^2-6uc+c^2)ab(u^3+3u^2c+3uc^2+c^3)\right)=$$ $$=-(u+v)^2C\left(\sum_{cyc}ab(c^5-6uc^4+3uc^4)\right)=-(u+v)^2C\left(\sum_{cyc}w^3(a^4-3ua^3)\right)=$$ $$=-(u+v)^2(12u-9u)=-3u(u+v)^2<0,$$ which says that $f$ is a concave function, which gives that it's enough to prove $f(w^3)\geq0$ for an extremal value of $w^3$, which by $uvw$ happens in the following cases.

1. $w^3=0$.

Let $c=0$ and $b=\frac{3}{a},$ where $a>0$.

Thus, after this substitution to $$8\left(\sum\limits_{cyc}(a^2+5ab)\right)^3(a+b+c+abc)\geq(a+b+c+3)^2\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+4c)^3$$ we obtain: $$a^{12}-10a^{11}+68a^{10}-192a^9+1077a^8-2736a^7+5049a^6-8208a^5+$$ $$+9693a^4-5184a^3+5508a^2-2430a+729\geq0,$$ which is true because $$a^{12}-10a^{11}+68a^{10}-192a^9+1077a^8-2736a^7+5049a^6-8208a^5+$$ $$+9693a^4-5184a^3+5508a^2-2430a+729=$$ $$=\left(a^{12}-10a^{11}+68a^{10}-192a^9+177a^8\right)+$$ $$+\left(900a^8-2736a^7+5049a^6-8208a^5+5652a^4\right)+$$ $$+\left(4041a^4-5184a^3+5508a^2-2430a+729\right)>0.$$ 2. Two variables are equal.

Let $b=a$ and $c=\frac{3-a^2}{2a},$ where $0<a\leq\sqrt3.$

Thus, we need to prove that: $$(a-1)^4\left(a^{12}+40^{11}+130a^{10}+392a^9+887a^8+1040a^7+2044a^6+656a^5+927a^4-56a^3+98a^2-24a+9\right)\geq0,$$ which is obvious.

Answer 3

How we can get $(a+b+4c)^3$ in the following Holder's step? $$\sum_{cyc}\sqrt{\frac{a+b}{ab+1}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{a+b}{ab+1}}\right)^2\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+4c)^3}{\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+4c)^3}}\geq$$ $$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(a+b)(a+b+4c)\right)^3}{\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+4c)^3}}=\sqrt{\frac{8\left(\sum\limits_{cyc}(a^2+5ab)\right)^3}{\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+4c)^3}}.$$ We need to prove $$\sum_{cyc}\sqrt{\frac{a+b}{ab+1}}\geq\frac{a+b+c+3}{\sqrt{a+b+c+abc}},$$ where $a$, $b$ and $c$ are non-negatives such that $ab+ac+bc=3$.

Let $c=0$ and $a=b=\sqrt3$.

Thus, should be $$2\sqrt[4]3+\frac{\sqrt[4]{12}}{2}-\frac{2\sqrt3+3}{\sqrt[4]{12}}>0.$$ But $$2\sqrt[4]3+\frac{\sqrt[4]{12}}{2}-\frac{2\sqrt3+3}{\sqrt[4]{12}}=0.0896...$$ Now,for $k\geq0$ by Holder $$\sum_{cyc}\sqrt{\frac{a+b}{ab+1}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{a+b}{ab+1}}\right)^2\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+kc)^3}{\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+kc)^3}}\geq$$ $$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(a+b)(a+b+kc)\right)^3}{\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+kc)^3}}=\sqrt{\frac{8\left(\sum\limits_{cyc}(a^2+(k+1)ab)\right)^3}{\sum\limits_{cyc}(a+b)^2(ab+1)(a+b+kc)^3}}.$$ Holder it's the following.

Let $a_i>0$, $b_i>0$,$\alpha>0$ and $\beta>0$. Prove that: $$\left(\sum_{i=1}^na_i\right)^{\alpha}\left(\sum_{i=1}^nb_i\right)^{\beta}\geq\left(\sum_{i=1}^n\left(a_i^{\alpha}b_i^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}.$$

The equality occurs, when $(a_1,a_2,...,a_n)||(b_1,b_2,...,b_n).$

Id est, in the previous Holder the equality occurs for $$\left(\sqrt{\frac{a+b}{ab+1}},\sqrt{\frac{b+c}{bc+1}},\sqrt{\frac{c+a}{ca+1}}\right)||$$ $$||\left((a+b)^2(ab+1)(a+b+kc)^3,(b+c)^2(bc+1)(b+c+ka)^3,(c+a)^2(ca+1)(c+a+kb)^3\right),$$ which gives $$\sqrt{(a+b)(ab+1)}(a+b+kc)=\sqrt{(b+c)(bc+1)}(b+c+ka)=\sqrt{(c+a)(ca+1)}(c+a+kb),$$ which for $c=0$ and $a=b=\sqrt3$ gives: $$\sqrt{2\sqrt3\cdot4}(2\sqrt3+0)=\sqrt{\sqrt3\cdot1}(\sqrt3+k\sqrt3)$$ or $$k=4\sqrt2-1\approx4.65685$$ and we can try to use $k=4$, which leads to a right inequality.