Problem. Let $a,b,c\ge 0:ab+bc+ca=1.$ Prove that $$\frac{1}{\sqrt{a+bc}}+\frac{1}{\sqrt{b+ca}}+\frac{1}{\sqrt{c+ab}}\ge 2\sqrt{2}.$$ Equality holds when two of $a,b,c$ very near $0$.

The form of problem is quite similar with Pham Kim Hung's inequality (AOPSlink) and a problem I have posted recently (similarMSE)

In the post #46, Michael Rozenberg (aka arqady) gave a very sharp Holder estimate.

They are two same equality case. That why I tried to use similar Holder approach: \begin{align*} \left(\sum_{cyc}\frac{1}{\sqrt{a+bc}}\right)^2\sum_{cyc}(a+bc)(b+c)^3(3a+b+c)^3&\geq\left(\sum_{cyc}(b+c)(3a+b+c)\right)^3\\&=8\left(\sum_{cyc}(a^2+4ab)\right)^3 \end{align*}Hence, we need to prove:$$\left(\sum_{cyc}(a^2+4ab)\right)^3\geq\sum_{cyc}(a+bc)(b+c)^3(3a+b+c)^3,$$hay$$\left((a+b+c)^2+2\right)^3$$ $$ \geq (a+bc)(b+c)^3(3a+b+c)^3+(b+ca)(a+c)^3(3b+a+c)^3+(c+ba)(b+a)^3(3c+b+a)^3,$$ which is wrong when $a=b\rightarrow 0;c\rightarrow +\infty !$

I am trying to find a good Holder estimate.

If you find something interesting in using Holder (like tricks, the motivations...) please share me. Thank you.

  • $\begingroup$ Lagrange Mutliper is the quickest in such situations. $\endgroup$ Jun 17, 2023 at 1:04
  • $\begingroup$ Could you tell more detail ? $\endgroup$
    – TATA box
    Jun 17, 2023 at 1:39
  • $\begingroup$ Well keep track on my website, I will give many solved examples of Lagrange. $\endgroup$ Jun 17, 2023 at 1:45
  • $\begingroup$ Please send me link. Thank you. $\endgroup$
    – TATA box
    Jun 17, 2023 at 4:11
  • $\begingroup$ For sure, I will do. $\endgroup$ Jun 17, 2023 at 4:23

2 Answers 2


I hope the following will help.

By AM-GM $$\sqrt{2(a+bc)}\le a+bc+\frac{1}{2},$$hence we will prove$$\frac{1}{2a+2bc+1}+\frac{1}{2b+2ca+1}+\frac{1}{2c+2ab+1}\ge 1.$$ The last inequality was here.


I hope, the following will help.

By AM-GM $$\sum_{cyc}\frac{1}{\sqrt{a+bc}}=\sqrt{\sum_{cyc}\left(\frac{1}{a+bc}+\frac{2}{\sqrt{(a+bc)(b+ac)}}\right)}\geq$$ $$\geq\sqrt{\sum_{cyc}\left(\frac{1}{a+bc}+\frac{4}{a+bc+b+ac}\right)}=\sqrt{\sum_{cyc}\left(\frac{1}{a+bc}+\frac{4}{(a+b)(1+c)}\right)}$$ and it's enough to prove that $$\sum_{cyc}\left(\frac{1}{a+bc}+\frac{4}{(a+b)(1+c)}\right)\geq8.$$ If we'll prove that it's enough to prove the last inequality in two cases:

  1. $abc=0$;

  2. Two variables are equal,

so we can end a proof.

Indeed, for $c=0$ and $ab=1$ by AM-GM we obtain: $$\sum_{cyc}\left(\frac{1}{a+bc}+\frac{4}{(a+b)(1+c)}\right)=$$ $$=\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}+4\left(\frac{1}{a+b}+\frac{1}{a(1+b)}+\frac{1}{b(1+a)}\right)=$$ $$=a+b+1+4\left(\frac{1}{a+b}+\frac{1}{a+1}+\frac{1}{b+1}\right)=$$ $$=a+b+\frac{4}{a+b}+1+4\cdot\frac{2+a+b}{2+a+b}\geq4+1+4=9>8.$$ Also, for $b=a$ and $c=\frac{1-a^2}{2a},$ where $0<a<1$ we need to prove that $$a(8a^7-12a^6-21a^5+41a^4-14a^3-10a^2+15a+1)\geq0,$$ which is true because $$8a^7-12a^6-21a^5+41a^4-14a^3-10a^2+15a+1=$$ $$=a^4(8a^3-12a^2-21a+32)+\left(3a^2-\frac{7}{3}a+\frac{1}{3}\right)^2+\frac{(1-a)(157a+8)}{9}\geq0.$$

  • 1
    $\begingroup$ What a beautiful solution! Thank you very much @Michael Rozenberg. It is just a little typo in first line $\frac{1}{\sqrt{a+bc}}$. $\endgroup$
    – TATA box
    Jun 17, 2023 at 13:47
  • $\begingroup$ @Trần Nk Trang Thank you, I fixed. Now, we need to prove that it's enough to check two cases about them I said. $\endgroup$ Jun 17, 2023 at 13:50
  • $\begingroup$ We see that it is equivalent to $$f(w^3)=kw^6+A(u,v^2)w^3+B(u,v^2)\ge 0$$where $k<0$. Am I in right way ? $\endgroup$
    – TATA box
    Jun 17, 2023 at 14:26
  • $\begingroup$ @Trần Nk Trang I can't got it. Maybe I am wrong. I'll check it again later. $\endgroup$ Jun 17, 2023 at 14:38
  • 1
    $\begingroup$ sorry I made a mistake. $\endgroup$
    – TATA box
    Jun 19, 2023 at 5:33

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