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Let $a,b,c\ge 0: ab+bc+ca>0$ and $a+b+c+abc=4.$ Prove$$\sqrt{\frac{ab+9}{ab+9c}}+\sqrt{\frac{bc+9}{bc+9a}}+\sqrt{\frac{ca+9}{ca+9b}}\ge 3.$$

I tried to used AM-GM as follow:$$LHS\ge 3\sqrt[6]{\frac{(ab+9)(bc+9)(ca+9)}{(ab+9c)(bc+9a)(ca+9b)}},$$and we need to prove $(ab+9)(bc+9)(ca+9)\ge (ab+9c)(bc+9a)(ca+9b)$ but it is wrong when $a=b=2;c=0.$

Also by AM-GM$$\sum_{cyc}\sqrt{\frac{ab+9}{ab+9c}}=\sum_{cyc}\frac{ab+9}{\sqrt{(ab+9)(ab+9c)}}\ge 2\sum_{cyc}\frac{ab+9}{2ab+9c+9},$$and we need to prove $\sum_{cyc}\dfrac{ab+9}{2ab+9c+9}\ge \dfrac{3}{2},$ which is also wrong when $a=b=2;c=0.$

I think we can use Holder as$$\left(\sum_{cyc}\sqrt{\frac{ab+9}{ab+9c}}\right)^2.\sum_{cyc}(ab+9c)(ab+9)^2(kc+a+b)^3\ge \left(\sum_{cyc}(kc+a+b)(ab+9)\right)^3 .$$The last inequality is $$\left(\sum_{cyc}(kc+a+b)(ab+9)\right)^3 \ge 9\sum_{cyc}(ab+9c)(ab+9)^2(kc+a+b)^3.$$I check $k=0$ which is not good enough.

Does mixing variables method work? I hope you can help me prove this inequality. Thank you very much.

TATA box
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2 Answers2

1

If you allow computer-based proofs, here is such a proof done with Mathematica 13.3.1:

FindInstance[{p + r + s < 3 && p^2 == (a*b + 9)/(a*b + 9 c) && 
r^2 == (c*b + 9)/(c*b + 9 a) && s^2 == (a*c + 9)/(a*c + 9 b) && 
a + b + c + a*b*c == 4 && {a, b, c, p, r, s} >= 0}, {a, b, c, p, r,s}, Reals]

{}

FindInstance[{p + r + s == 3 && p^2 == (a*b + 9)/(a*b + 9 c) && 
r^2 == (c*b + 9)/(c*b + 9 a) && s^2 == (a*c + 9)/(a*c + 9 b) && 
a + b + c + a*b*c == 4 && {a, b, c, p, r, s} >= 0}, {a, b, c, p, r,s}, Reals]

{{a -> 1, b -> 1, c -> 1, p -> 1, r -> 1, s -> 1}}

user64494
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1

Some thoughts.

Fact 1. Let $x, y, z\ge 0$ with $x^2y^2 + y^2z^2 + z^2x^2 + \frac52 x^2 y^2 z^2 \ge \frac{11}{2}$. Then $x + y + z \ge 3$.
(The proof is given at the end.)

Let $$x := \sqrt{\frac{ab+9}{ab+9c}}, \quad y := \sqrt{\frac{bc+9}{bc+9a}}, \quad z := \sqrt{\frac{ca+9}{ca+9b}}.$$

We have $$x^2y^2 + y^2z^2 + z^2x^2 + \frac52 x^2 y^2 z^2 - \frac{11}{2} \ge 0. \tag{1}$$ (1) is verified by Mathematica.

By Fact 1, we have $x + y + z \ge 3$.

$\phantom{2}$

Proof of Fact 1.

Equivalently, we need to prove that, for all $x, y, z\ge 0$ with $x+y+z < 3$, $$x^2y^2 + y^2z^2 + z^2x^2 + \frac52x^2y^2z^2 < \frac{11}{2}. \tag{A1}$$

WLOG, assume that $x + y < 2$. It suffices to prove that, for all $x, y \ge 0$ with $x + y < 2$, $$x^2y^2 + y^2(3 - x - y)^2 + (3 - x - y)^2x^2 + \frac52x^2y^2(3 - x - y)^2 < \frac{11}{2}. \tag{A2}$$

Let $p = x + y, q = xy$. We have $0 \le p < 2$ and $p^2 \ge 4q$.

(A2) is written as $$f(q) := -\frac{5(p-3)^2 + 2}{2}q^2 + (2p^2 - 12p + 18)q - 9p^2 - p^4 + 6p^3 + \frac{11}{2} > 0.$$

Note that $f(q)$ is concave. Also, we have $f(0) = - 9p^2 - p^4 + 6p^3 + \frac{11}{2} > 0$ and $$f(p^2/4) = \frac{-5p^4 + 10p^3 - 3p^2 + 44p + 44}{32}(2 - p)^2 > 0.$$ Thus, $f(q) > 0$ on $[0, p^2/4]$. We are done.

River Li
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  • Nice solution again! – TATA box Oct 27 '23 at 15:30
  • @TATAbox I hope the pqr works well for (1). – River Li Oct 27 '23 at 15:32
  • Uvw is also a good way – TATA box Oct 27 '23 at 23:36
  • @TATAbox By the way, usually we want to see a human verifiable proof so that we can check every steps of the proof as usual. We can use computer (CAS, computer algebra system) to verify the inequalities. Usually, I am much interested in a proof which can be verified by hand easily (it can be very difficult to motivate the proof by hand). – River Li Oct 28 '23 at 00:35
  • @TATAbox Here is an example. I give a computer-assisted proof. It is difficult to verify by hand (perhaps 10000 hours). So it is not a proof I want as usual. – River Li Oct 28 '23 at 00:53