If $ab+bc+ca=1,$ prove $\frac{\sqrt{a+1}}{a+bc}+\frac{\sqrt{b+1}}{b+ca}+\frac{\sqrt{c+1}}{c+ab}\ge 1+2\sqrt{2}.$

Question

Problem. Let $a,b,c$ be non-negative real numbers satisfying $ab+bc+ca=1.$ Prove that $$\frac{\sqrt{a+1}}{a+bc}+\frac{\sqrt{b+1}}{b+ca}+\frac{\sqrt{c+1}}{c+ab}\ge 1+2\sqrt{2}.$$

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I've tried to use Holder inequality without success.

$$\left(\sum_{cyc}\frac{\sqrt{a+1}}{a+bc}\right)^2.\sum_{cyc}(a+1)^2(a+bc)^2\ge [a+b+c+3]^3. \tag{1}$$

$$\left(\sum_{cyc}\frac{\sqrt{a+1}}{a+bc}\right)^2.\sum_{cyc}(a+1)^2(a+bc)^2(b+c)^3\ge \left(\sum_{cyc}(a+1)(b+c)\right)^3.\tag{2} $$

Which are both leads to wrong inequalities in general.

I'd like to ask two questions.

1. Is there a better Holder using ?

I think the appropriate one might be ugly but if you find it, please feel free to share it here.

2. Are there others idea which are smooth enough?

For example, Mixing variables, AM-GM or Cauchy-Schwarz...etc.

I aslo hope to see a good lower bound of $\frac{\sqrt{a+1}}{a+bc},$ which eliminates the radical form (may be the rest is simpler by $uvw$)

All ideas and comments are welcomed. Thank you for your interest!

Remark. About $uvw$, see here.

Answer 1

Some thoughts.

WLOG, assume that $a \ge b \ge c$.

By AM-GM, we have $\sqrt{a+1} = \frac{2(a + 1)\sqrt 2}{2\sqrt{a + 1}\sqrt{2}} \ge \frac{2(a + 1)\sqrt 2}{(a + 1) + 2}$. Also, we have $(1 + c) - (1 + c/3)^2 = \frac19 c(3 - c)\ge 0$ which results in $\sqrt{1 + c} \ge 1 + c/3$. It suffices to prove that $$ \frac{2(a + 1)\sqrt 2}{(a + 3)(a + bc)} + \frac{2(b + 1)\sqrt 2}{(b + 3)(b + ca)} + \frac{1 + c/3}{c + ab} \ge 1 + 2\sqrt 2$$ or $$\frac{3 - 3ab - 2c}{3(c + ab)} \ge \left(2 - \frac{2(a + 1)}{(a + 3)(a + bc)} - \frac{2(b + 1)}{(b + 3)(b + ca)}\right)\sqrt 2. \tag{1}$$

Since $3 - 3ab - 2c = 1 - ab + (2 - 2c) > 0$, we only need to prove the case that $\mathrm{RHS}_{(1)} > 0$. Since $\sqrt 2 < \frac{10}{7}$, it suffices to prove that $$\frac{3 - 3ab - 2c}{3(c + ab)} \ge \left(2 - \frac{2(a + 1)}{(a + 3)(a + bc)} - \frac{2(b + 1)}{(b + 3)(b + ca)}\right)\cdot \frac{10}{7}. \tag{2}$$ (2) is true which can be verified by Mathematica.

Answer 2

Some thoughts.

Remark: All results are verified by Mathematica.

Fact 1. Let $x, y, z \ge 0$ with $x^2 + y^2 + z^2 \ge 5$ and $x^2y^2 + y^2z^2 + z^2x^2 + \frac32x^2y^2z^2 \ge 14$. Then $x + y + z \ge 1 + 2\sqrt 2$.

Let $$x = \frac{\sqrt{a + 1}}{a + bc}, \quad y = \frac{\sqrt{b + 1}}{b + ca}, \quad z = \frac{\sqrt{c + 1}}{c + ab}.$$

We have $x^2 + y^2 + z^2 \ge 5$ and $x^2y^2 + y^2z^2 + z^2x^2 + \frac32x^2y^2z^2 \ge 14$.

By Fact 1, we have $x + y + z \ge 1 + 2\sqrt 2$.

Answer 3

I have written some partial results about this problem, I hope it helps someone to find the solution.

Without loss of generality, $a\geq b \geq c$. Next, since $ab+bc+ac=1$, then at least one of them is less than $1/\sqrt{3}$, i.e., $c\leq 1/\sqrt{3}$. Otherwise, $ab+bc+ac>1$.

Similarly, $ab\leq ab+bc+ac \leq 1$, and hence if $b>1$, then $ab>1$, which is a contradiction. Therefore, $b\leq 1$. Note also that as $bc\leq bc+ac+ab=1$,

\begin{equation} \frac{\sqrt{a+1}}{a+bc}\geq \frac{1}{\sqrt{a+1}} \end{equation} Finally, combining all the results, \begin{equation} \frac{\sqrt{a+1}}{a+bc}+\frac{\sqrt{b+1}}{b+ac}+\frac{\sqrt{c+1}}{c+ab}\geq \sqrt{\frac{\sqrt{3}}{1+\sqrt{3}}}+\frac{1}{\sqrt{2}} \end{equation}

Answer 4

Here is another approach.

By AM-GM, we have $$\frac{\sqrt{a+1}}{a+bc} = \frac{2(a+1)}{(a+bc)\cdot 2\sqrt{a + 1}} \ge \frac{2(a+1)}{(a+bc)\cdot \left(\frac{a + 1}{(\sqrt 2 - 1)a + 1} + (\sqrt 2 - 1)a + 1\right)}.$$

It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{2(a+1)}{(a+bc)\cdot \left(\frac{a + 1}{(\sqrt 2 - 1)a + 1} + (\sqrt 2 - 1)a + 1\right)} \ge 1 + 2\sqrt 2. \tag{1}$$ (1) is true which is verified by Mathematica.

Answer 5

Just an idea :

Transforming the sum into a product :

Let $c\in (0,1)$ define :

$$a=\frac{\left(1-xc\right)}{x+c},b=x$$

Then using AM-GM setting $d=\left|\frac{1}{\left(1-c\right)c}\right|$ it seems we have :

$$3\left(\left(\frac{\sqrt{a+1}}{a+bc}+\frac{d}{3}\right)\left(\frac{\sqrt{c+1}}{c+ab}+\frac{d}{3}\right)\left(\frac{\sqrt{b+1}}{b+ca}+\frac{d}{3}\right)\right)^{\frac{1}{3}}-d-\left(1+2\sqrt{2}\right)>0$$

Can someone confirm ?

Edit : the inequality is false like that but we can replace $d=\left|\frac{1}{\left(1-c\right)c}\right|$ by $d=\frac{1}{\left(1-c\right)^2c^2}$

Now I think it's true give me feed back

We have a really nice bound which need to be simplify let $x\geq 0$ then we have :

$$f\left(x\right)=\frac{\left(1+\left(\frac{51}{2\sqrt{2}}-18\right)\right)\left(x+1\right)}{2+\frac{x}{6}-\frac{1}{\frac{x}{3}+1}}-\left(\frac{51}{2\sqrt{2}}-18\right)\left(1+\frac{x}{3}\right)-b\frac{x}{\left(x+3\right)^{3}}\left(\frac{2x}{x+1}-1\right)^{2}\left(3-x\right)-\sqrt{x+1}\leq 0$$

$b$ is defined as follow :

$$f'(3)=0,b\simeq 3/4$$

Answer 6

Just an idea...

Let \begin{equation} f(a,b,c)=\frac{\sqrt{a+1}}{a+bc}+\frac{\sqrt{b+1}}{b+ac}+\frac{\sqrt{c+1}}{c+ab} \end{equation} Then $\frac{\partial f}{\partial c}<0$ gives $ab<1+c$ which is true. Similarly for partial derivatives with respect to $a$ and $b$. So the function is decreasing away from boundry. Without loss of generality, we can take $c=0$. Rest should be easy. We have two variables.