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

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

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.

1. 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.

• I checked (1) is true Oct 9, 2023 at 1:57
• @above How about is the result when a=b-> 0? Oct 9, 2023 at 11:04
• Oct 12, 2023 at 14:39
• The idea with $x^2+y^2+z^2+k\sum\limits_{cyc}x^2y^2+mx^2y^2z^2\geq5+8k+4m$ does not help. Oct 13, 2023 at 8:22
• @MichaelRozenberg Do you use Holder inequality? Oct 13, 2023 at 10:31

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 $$x + y + z \ge 1 + 2\sqrt 2$$.

• This way is beautiful. We can use contradiction method and Schur to prove the fact. Oct 15, 2023 at 0:23
• I tried to use AM-GM: $$\frac{\sqrt{a+1}}{a+bc}=\frac{a+1}{a+bc}.\frac{1}{\sqrt{a+1}}.$$Now, we'll find suitable yield $2\sqrt{a+1}.f(a,b,c)\le a+1+f^2(a,b,c),$ which savesoccuring equality. I failed here. Oct 30, 2023 at 10:40

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.

• It's interesting. Thank you. Oct 14, 2023 at 7:53
• @RiverLi we have a nice bound ie $$\sqrt{x+1}\geq \frac{x+1}{\frac{x}{6}+2-\frac{1}{\frac{x}{3}+1}},\forall x\ge 0$$ Oct 16, 2023 at 9:06
• @ErikSatie Yes, it is good to find a uniform bound. You can write down your proof. Oct 16, 2023 at 9:26
• Well it's very ugly now I have : $$\sqrt{x+1}\geq \left(1+\frac{51}{2\sqrt{2}}-18\right)\frac{\left(x+1\right)}{2+\frac{x}{6}-\frac{1}{\frac{x}{3}+1}}-\left(1+\frac{x}{3}\right)\left(\frac{51}{2\sqrt{2}}-18\right)-\frac{\left|x\left(x-1\right)\left(x-3\right)\right|}{\left(x+3\right)^{3}},\forall x\geq 0$$ which is too complicated for a proof .But it's uniform . Oct 16, 2023 at 10:27
• @ErikSatie When applying your bound for the inequality, I checked and it seems not true. Did you check it? Oct 16, 2023 at 11:41

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 $$bc\leq bc+ac+ab=1$$,

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

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.

• Oh, you wrote my idea as an answer. Very nice, thank you. Oct 30, 2023 at 10:41
• @TATAbox Yes, the idea is similar to some problems. Oct 30, 2023 at 11:28

Just an idea...

Let $$$$f(a,b,c)=\frac{\sqrt{a+1}}{a+bc}+\frac{\sqrt{b+1}}{b+ac}+\frac{\sqrt{c+1}}{c+ab}$$$$ Then $$\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.

• But there is a constraint $ab+bc+ca=1$. You cannot (e.g.) decrease $c$ while keeping the other variables constant. Oct 16, 2023 at 11:47
• @MartinR I thought without the constraint first then restricted on the constraint surface. My advanced calculus is not enough here. Oct 16, 2023 at 20:39

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 :

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