Let $a,b,c\ge 0: ab+bc+ca>0$ and $a+b+c+abc=4.$ Prove that$$\color{black}{\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ca}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}. }$$


My teacher asasigned this problem to our class as a homework. I post it here to look for help and share some thoughts.

Any ideas and comments are welcome. Please feel free to discuss about this inequality.

Here is my attempts.

Since equality holds at $a=b=2;c=0$ we can't use normal approach.

For example, by using Cauchy-Schwarz inequality$$\color{black}{\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{9}{\sqrt{a^2+4bc}+\sqrt{b^2+4ba}+\sqrt{c^2+4ba}}. }$$ But $$\frac{9}{\sqrt{a^2+4bc}+\sqrt{b^2+4ba}+\sqrt{c^2+4ba}}- \frac{5}{4}=-\frac{1}{8}$$is already wrong when $a=b=2;c=0.$

I'll post more ideas which are not good enough. Now, I hope you consider carefully before voting to close my topic.

I try to use Jichen lemma which seems not good enough.

Indeed, we can rewrite the original inequality as $$\color{black}{\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{1}{\sqrt{16}}+\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}. }$$ I checked that $$\frac{1}{a^2+4bc}+\frac{1}{b^2+4ca}+\frac{1}{c^2+4ab}\ge \frac{9}{16}$$ is not true when $a=b=0.5;c=2.4$.

Also, in comment section, Michael Rozenberg said that the Holder using with $(3a+b+c)^3$ is not good. I hope you can optimize your idea soon.

I'll update more approach when I found it. Thanks for your interest.

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    $\begingroup$ Can someone, which want to close this topic ,explain us, why did you do it? $\endgroup$ Nov 27, 2023 at 7:42
  • 1
    $\begingroup$ I don't agree. Your problem is very nice. $\endgroup$ Nov 27, 2023 at 8:14
  • 1
    $\begingroup$ Thank you for your encouragement. $\endgroup$
    – Dragon boy
    Nov 27, 2023 at 8:20
  • 6
    $\begingroup$ It's really not your fault (or that of your problem). I'm afraid Math.SE has seen an increase in drive-by down-voters and close-voters recently. I suppose they think it's a sign of greater discrimination, when really it's just conceit. If a (non-PSQ) problem is deficient enough for down-votes and close-votes, it's deficient enough for a comment as to why. But go talk to a brick wall. $\endgroup$
    – Brian Tung
    Nov 30, 2023 at 5:33
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    $\begingroup$ Not only that, but some of them have shockingly poor taste. I've seen close-votes and down-votes the last few days on really interesting questions, with a reasonable attempt at solving them shown in the post. $\endgroup$
    – Brian Tung
    Nov 30, 2023 at 5:36

4 Answers 4


Some thoughts.

By Holder inequality, we have \begin{align*} &\left(\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a^2 + 4bc}} \right)^2\cdot \sum_{\mathrm{cyc}} (a^2 + 4bc)(4b + 4c - bc + 4ab + 4ac)^3 \\ \ge{}& \left(\sum_{\mathrm{cyc}} (4b + 4c - bc + 4ab + 4ac)\right)^3. \tag{1} \end{align*}

It suffices to prove that \begin{align*} &\left(\sum_{\mathrm{cyc}} (4b + 4c - bc + 4ab + 4ac)\right)^3 \\[6pt] \ge{}& \left(\frac54\right)^2 \sum_{\mathrm{cyc}} (a^2 + 4bc)(4b + 4c - bc + 4ab + 4ac)^3. \tag{2} \end{align*}

(2) is true which is verified by Mathematica.

  • $\begingroup$ Very nice Holder using! +1 $\endgroup$
    – TATA box
    Nov 29, 2023 at 12:24
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    $\begingroup$ @TATAbox My two proofs are complicated. According to OP's statement "My teacher asasigned this problem to our class as a homework." There should be a simple solution. $\endgroup$
    – River Li
    Nov 29, 2023 at 12:47
  • $\begingroup$ @RiverLi Do you have pqr solution? $\endgroup$
    – Dragon boy
    Dec 1, 2023 at 5:30
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    $\begingroup$ @Dragonboy I can not find a simple proof in my two answers (complicated proofs are not what I want). I give up currently. Hope to see better solutions. $\endgroup$
    – River Li
    Dec 1, 2023 at 14:52

Disclaimer: Not a full solution

If one of $a, b, c=0$:

WLOG let $a=0$. Since $ab+bc+ca>0$ we know that $b, c\neq0$.

$$\begin{align} & \frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ca}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4} \\ \iff &\frac{1}{\sqrt{4bc}}+\frac{1}{b}+\frac{1}{c}\geq\frac{5}{4} \\ \iff &\frac{2\sqrt{bc}+4b+4c}{4bc}\geq\frac{5}{4} \\ \iff &8\sqrt{bc}+16b+16c\geq20bc \\ \iff &8\sqrt{bc}+64\geq20bc &\tag{1}\\ \iff &8x+64\geq20x^2 \tag{2}\\ \iff &20x^2-8x-64\leq0 \end{align}$$

$(1)$: The condition $a+b+c+abc=4$ becomes $b+c=4$.

$(2)$: Here $x=\sqrt{bc}$.

However by AM-GM, $0\leq x\leq\dfrac{b+c}{2}=2$. Since $2$ is the bigger root of $20x^2-8x-64$, we know that $20x^2-8x-64\leq0$ is true. Hence, the statement is proved.

If $a, b, c\neq0$:

$$\begin{align} &\sum\frac{1}{\sqrt{a^2+4bc}}\geq\frac{5}{4} \\ \iff &\sum\frac{\sqrt{a^2+4bc}}{a^2+4bc}\geq\frac{5}{4} \\ \iff &4\sum\sqrt{a^2+4bc}(b^2+4ca)(c^2+4ab)\geq5(a^2+4bc)(b^2+4ca)(c^2+4ab) \end{align}$$

Let $x=a^2+4bc$, $y=b^2+4ca$, $z=c^2+4ab$.

$$\begin{align} \iff &4\sum xy^2z^2\geq 5x^2y^2z^2 \\ \iff &4xy+4yz+4zx\geq 5xyz \end{align}$$

I need to go now, but I will continue from here if I have the time. I have a feeling that the final step is Schur's Inequality and maybe some of you can try that. Anyone who manages to solve the question from my works, please give me credit.


Some thoughts.

By AM-GM, it suffices to prove that $$\frac{2}{\frac{a^2 + 4bc}{2 + bc/2} + (2 + bc/2)} + \frac{2}{\frac{b^2 + 4ca}{2 + ca/2} + (2 + ca/2)} + \frac{2}{\frac{c^2 + 4ab}{2 + ab/2} + (2 + ab/2)} \ge \frac54. \tag{1}$$

(1) is true which is verified by Mathematica.


Remark 1: This is my third proof (the idea) which is better than my two old proofs.

Remark 2: Some years ago, I used a similar idea for the problem: Let $a, b, c > 0$ with $a + b + c = 3$. Prove that $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \ge 3$. I gave equivalently the following fact (then let $x = \sqrt[5]{\frac{a+1}{a+b}}$ etc.):

Fact 1: Let $x, y, z \ge 0$ such that $x^5y^5z^5 \ge \frac{25}{27}$ and $3x^5y^5z^5 - 2(x^5y^5 + y^5z^5 + z^5x^5) + 2(x^5 + y^5 + z^5) = 3$. Then $x^2 + y^2 + z^2 \ge 3$.

I also used a similar trick in P1.

Remark 3: I think that the spirit of this trick is similar to the pqr method.

  • In the pqr method, we use the substitution $p = a + b + c, q = ab + bc + ca, r = abc$. The desired inequality is written in terms of $p, q, r$, say $f(p, q, r) \ge 0$. Then we seek some relations among $p, q, r$ which can implies $f(p, q, r) \ge 0$. The relation among $p, q, r$ are $p^2 \ge 3q, p^3 - 4pq + 9r \ge 0$, $q^2 \ge 3pr$ etc.

  • In this trick, we use $x, y, z$ to substitute the radicals, and the desired inequality is written in terms of $x, y, z$, say$f(x, y, z)\ge 0$. Then we seek some relations among $x^2, y^2, z^2$ which can implies $f(x, y, z) \ge 0$. The relations among $x^2, y^2, z^2$ clearly contain no radicals, sometimes it is easy to handle.

Remark 4: I used this trick recently in P2, P3, P4, P5.

Some thoughts.

Fact 1: Let $x, y, z > 0$ with $x^2 + y^2 + z^2 \le 24$ and $(x^2 + y^2 + z^2)(x^2y^2 + y^2z^2 + z^2x^2) - 24x^2y^2z^2 + 2688 \ge 0$. Then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \ge \frac54$.
(It is verified by Mathematica. I think the pqr method works. )

Now, let $$x = \sqrt{a^2 + 4bc}, \quad y = \sqrt{b^2 + 4ca}, \quad z = \sqrt{c^2 + 4ab}.$$

We can prove that $$x^2 + y^2 + z^2 = a^2 + b^2 + c^2 + 4ab + 4bc + 4ca \le 24, \tag{1}$$ and $$(x^2 + y^2 + z^2)(x^2y^2 + y^2z^2 + z^2x^2) - 24x^2y^2z^2 + 2688 \ge 0. \tag{2}$$ (1) and (2) are both true which are verified by Mathematica. The proof of (1) is easy. I think the pqr method works for (2): Let $p = a + b + c, q = ab + bc + ca, r = abc$. (2) is written as $$4\,{p}^{4}q-374\,{p}^{3}r+{p}^{2}{q}^{2}+1460\,pqr-110\,{q}^{3}-3000\, {r}^{2}+2688 \ge 0.$$

By Fact 1, we have $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \ge \frac54$.

  • $\begingroup$ Very useful answer! Thank you again +1. $\endgroup$
    – Dragon boy
    Dec 2, 2023 at 1:18
  • $\begingroup$ @Dragonboy You are welcome. $\endgroup$
    – River Li
    Dec 2, 2023 at 1:20
  • $\begingroup$ @RiverLi Could you please tell me how did you come up with the fact 1? $\endgroup$
    – Anonymous
    Dec 2, 2023 at 2:00
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    $\begingroup$ @Anonymous I have no systematic approach. I do it with try and error. $\endgroup$
    – River Li
    Dec 2, 2023 at 2:05
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    $\begingroup$ @RiverLi Bravo for your nice try. $\endgroup$
    – Anonymous
    Dec 2, 2023 at 3:03

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