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.
- 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.
- 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.
Best Answer
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 $x + y + z \ge 1 + 2\sqrt 2$.