Problem. Let $a,b,c>0$, prove that $$\frac{1}{\sqrt[3]{abc}}+5\ge 2\sqrt{3}\left(\frac{1}{\sqrt{a+2}}+\frac{1}{\sqrt{b+2}}+\frac{1}{\sqrt{c+2}}\right).$$

I have an ugly proof. After replacing $(a,b,c)\rightarrow (x^3,y^3,z^3)$, we $$ \frac{1}{xyz}+5\ge 2\sum_{\mathrm{cyc}}{\frac{1}{\sqrt{\frac{x^3+2}{3}}}}.$$ And I use $$\frac{1}{\sqrt{\frac{x^3+2}{3}}}\le \frac{4(x+17)}{29x^2-18x+61},$$ which is true after squaring checking.

It remains to prove $$ \frac{1}{xyz}+5\ge 2\sum_{\mathrm{cyc}}{\frac{4(x+17)}{29x^2-18x+61}}.$$I verified by Maple, which is very ugly but it is true.

I hope we can find some simple proofs. Thank you.

  • $\begingroup$ Just a reminder that this comes under the tag contest-math too. $\endgroup$
    – IraeVid
    Jun 22 at 13:55
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    $\begingroup$ What does it mean ? $\endgroup$ Jun 22 at 14:28
  • $\begingroup$ I see that most of your questions have got answers. Please accept one answer that you believe is the best solution to your problem. $\endgroup$
    – Akira
    Jun 24 at 12:50
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    $\begingroup$ Ok, thank you for reminding me $\endgroup$ Jun 24 at 13:13
  • $\begingroup$ Very hard problem. Is there classical application ? $\endgroup$
    – TATA box
    Aug 2 at 7:09

2 Answers 2


Easy to see that for $c\rightarrow0^+$ or for $c\rightarrow+\infty$ the inequality holds.

Let $$f(a,b,c)=\frac{1}{\sqrt[3]{abc}}+5-2\sqrt3\sum_{cyc}\frac{1}{\sqrt{a+2}}.$$ Thus, in the inside minimum point should be $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0.$$ Let in this point $a\neq b$, $a\neq c$ and $b\neq c$.

Thus, $$\frac{1}{3\sqrt{a^4bc}}-\frac{\sqrt3}{\sqrt{(a+2)^3}}=\frac{1}{3\sqrt{ab^4c}}-\frac{\sqrt3}{\sqrt{(b+2)^3}}=\frac{1}{3\sqrt{abc^4}}-\frac{\sqrt3}{\sqrt{(c+2)^3}}=0,$$ which gives $$\frac{1}{3\sqrt3\sqrt[3]{abc}}=\frac{a}{\sqrt{(a+2)^3}}=\frac{b}{\sqrt{(b+2)^3}}=\frac{c}{\sqrt{(c+2)^3}}$$ and from here $$\frac{a^2}{(a+2)^3}=\frac{b^2}{(b+2)^3}$$ or $$(a-b)(a^2b^2-12ab-8a-8b)=0,$$ which by AM-GM gives $$a^2b^2=12ab+8a+8b\geq12ab+16\sqrt{ab},$$ which gives $ab\geq16$ and by the same way $ac\geq16$ and $bc\geq16.$

Also, we have $$a^2b^2-a^2c^2=12ab+8a+8b-12ac-8a-8c$$ or $$(b-c)(a^2(b+c)-12a-8)=0$$ or $$12a+8=a^2(b+c).$$ By the same way we obtain also $$12b+8=b^2(a+c)$$ and from here $$(a-b)(ab+ac+bc-12)=0,$$ which gives $$ab+ac+bc=12,$$ which is a contradiction because $$12=ab+ac+bc\geq3\cdot16\geq48.$$ Thus, in the minimum point $(a,b,c)$ two numbers should be equal.

Let $b=a$ and $c\neq a$.

Thus, since $$(a-c)(a^2c^2-12ac-8a-8c)=0,$$ we obtain $$a^2c^2-12ac-8a-8c=0$$ or $$a^2c^2-4(3a+2)c-8a=0$$ or $$c=\frac{2(3a+2)+\sqrt{4(3a+2)^2+8a^3}}{a^2}$$ or $$c=\frac{2(3a+2)+2(a+2)\sqrt{2a+1}}{a^2}$$ and since $$\frac{1}{\sqrt[3]{a^2c}}=\frac{3\sqrt3a}{\sqrt{(a+2)^3}},$$ it's enough to prove in this case that: $$\frac{3\sqrt3a}{\sqrt{(a+2)^3}}+5\geq2\sqrt3\left(\frac{2}{\sqrt{a+2}}+\frac{1}{\sqrt{\frac{2(3a+2)+2(a+2)\sqrt{2a+1}}{a^2}+2}}\right)$$ or $$\frac{3\sqrt3a}{a+2}+5\sqrt{a+2}\geq2\sqrt3\left(2+\frac{a}{\sqrt{2(a+1+\sqrt{2a+1})}}\right)$$ or $$\frac{3\sqrt3a}{a+2}+5\sqrt{a+2}\geq2\sqrt3\left(2+\frac{a}{\sqrt{2a+1}+1}\right)$$ or $$\frac{3\sqrt3a}{a+2}+5\sqrt{a+2}\geq\sqrt3\left(3+\sqrt{2a+1}\right)$$ or $$5\sqrt{a+2}\geq\sqrt3\left(\sqrt{2a+1}+\frac{6}{a+2}\right)$$ or $$19a^3+123a^2+264a+80\geq36(a+2)\sqrt{2a+1}$$ and since $$\sqrt{2a+1}\leq a+1,$$ it's enough to prove here that:$$19a^3+123a^2+264a+80\geq36(a+2)(a+1),$$ which is obvious.

For $a=c$ we need to prove that: $$\frac{1}{a}+5\geq\frac{6\sqrt3}{\sqrt{a+2}}$$ or $$(5a+1)^2(a+2)\geq108a^2,$$ which is true by AM-GM: $$(5a+1)^2(a+2)\geq\left(6\sqrt[6]{a^5}\right)^2\cdot3\sqrt[3]a=108a^2$$ and we are done!

  • $\begingroup$ Many thanks! I will search more to understand well. $\endgroup$ Jun 22 at 14:26
  • $\begingroup$ "But here I stopped", could you tell more? I think we should continue to point a contradiction and finally, we get a=b=c $\endgroup$ Jun 22 at 14:41
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    $\begingroup$ @Trần Nk Trang Yes, you are right, we can kill it now. $\endgroup$ Jun 23 at 4:40
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    $\begingroup$ I read your updated solution. It is great, Micheal Rozenberg ! $\endgroup$ Jun 23 at 11:17

Partial simple answer :

Let :


Then :

$$f''(x)\leq 0,\forall x\in[-\ln(4),\infty)$$

Then using Jensen's inequality we have $x\in[\frac{1}{4}\max\left(a,b,c\right),\infty]$ and $b\to b/x,a\to a/x,c\to c/x$:

$$2\sqrt{3}\left(\frac{1}{\sqrt{\frac{a}{x}+2}}+\frac{1}{\sqrt{\frac{b}{x}+2}}+\frac{1}{\sqrt{\frac{c}{x}+2}}\right)\leq 3g\left(\frac{x}{\left(abc\right)^{\frac{1}{3}}}\right),g\left(x\right)=2\sqrt{3}\left(\frac{1}{\sqrt{1/x+2}}\right)$$

Then I think you can finish that .

Edit for more details :

$$3*2\sqrt{3}\left(\frac{1}{\sqrt{\frac{\left(abc\right)^{\frac{1}{3}}}{x}+2}}\right)\leq 5+\frac{x}{\left(abc\right)^{\frac{1}{3}}}$$

Setting : $\frac{x}{\left(abc\right)^{\frac{1}{3}}}=1/y$ so :

$$3*2\sqrt{3}\left(\frac{1}{\sqrt{y+2}}\right)\leq 5+1/y$$

It's a second degree polynomial so think to push on both side the square .

Remark to have a complete proof :

We can use the intermediate inequality for $x\ge 0$ :

$$\frac{2\sqrt{3}}{\sqrt{x+2}}\leq f\left(x\right)=\frac{6}{\sqrt{x}+2}$$

Then use Jensen's inequality for $a,b\in[1,\infty)$ we have to show :

$$2f\left(\sqrt{ab}\right)+f\left(c\right)-5-\frac{1}{\left(abc\right)^{\frac{1}{3}}}\leq 0$$

Then we plugg $ab=u$ .

The inequality now is less harder .

  • $\begingroup$ Thank you, it seems hard to me. I am secondary student, I will try more. $\endgroup$ Jun 23 at 12:21
  • $\begingroup$ Ah so I shall give you more details and now I correct some mistake . $\endgroup$ Jun 23 at 18:32
  • $\begingroup$ I meant you used a strange term which is $e^$ $\endgroup$ Jun 24 at 0:51
  • $\begingroup$ This is the exponential .On the other hand the inequality is showed for $a,b,c\in(0,4]$ or $a,b,c\in[4,\infty)$ $\endgroup$ Jun 24 at 8:06
  • $\begingroup$ Have you thought about a elementary solution by AM-GM for this monster ? $\endgroup$ Jun 25 at 13:46

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