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For $a,b,c>0$ Prove that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$

My attempt: By AM-GM we obtain $$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge 3\sqrt[3]{\frac{a^2}{bc}}=\frac{3a}{\sqrt[3]{abc}}$$ Thus $$\sum \frac{a+c}{b}\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}$$ So it suffices to show that $$6\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}\Leftrightarrow 3\sqrt[3]{abc}\ge a+b+c$$ Which is clearly wrong. :"(

Thank you very much.

Michael Rozenberg
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BestChoice123
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    Nope, I tried it above. See – BestChoice123 Jul 05 '20 at 09:57
  • $\sum \frac{a+c}{b}\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}\Leftrightarrow \prod\left(1+\frac{a}{b}\right)\ge 2+\frac{2(a+b+c)}{\sqrt[3]{abc}}$ – BestChoice123 Jul 05 '20 at 10:00
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    I have seen but no ways can prove mine :( – BestChoice123 Jul 05 '20 at 10:11
  • @AlexeyBurdin Can you elaborate? Naively stringing together what you wrote, I get $ ( a + b+ c)( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ) + 3 = (1 + \frac{a}{b})(1+ \frac{b}{c})(1+ \frac{c}{a}) + 4 \geq 6 + 2 \frac{a+b+c} { \sqrt[3] { abc} }$. However, note that $ \frac{a+b+c} { \sqrt[3] {abc} } \geq 3 $, so I cannot complete it from here. – Calvin Lin Jul 05 '20 at 22:54
  • I was mistaken. The inequalities looks similar, but this is harder. @CalvinLin – Alexey Burdin Jul 06 '20 at 00:52
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    Why is this question being closed for being a duplicate? I went to the link that is claimed to be a duplicate, the inequality there is not the same as this one. I also don't see a quick way to convert one inequality into the other. – Batominovski Jul 06 '20 at 10:23
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    @Batominovski Context was that Alexey initially claimed it was a duplicate, in a now deleted comment with several upvotes. I pointed out that it wasn't immediately a duplicate, as you realized. (This is harder than the linked one). – Calvin Lin Jul 06 '20 at 23:52

3 Answers3

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Since $\prod\limits_{cyc}(a+b)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$, by AM-GM we obtain: $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3=\frac{\sum\limits_{cyc}(a^2b+a^2c+2abc)}{abc}=$$ $$=\frac{\sum\limits_{cyc}c(a+b)^2}{abc}\geq \frac{3\sqrt[3]{abc\prod\limits_{cyc}(a+b)^2}}{abc}\geq \frac{3\sqrt[3]{abc\cdot\frac{64}{81}(a+b+c)^2(ab+ac+bc)^2}}{abc}\geq$$ $$\geq \frac{3\sqrt[3]{abc\cdot\frac{64}{81}(a+b+c)^2\cdot3abc(a+b+c)}}{abc}=\frac{4(a+b+c)}{\sqrt[3]{abc}}.$$

Michael Rozenberg
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  • Good solution. In your 3rd line, it should be $\prod\limits_{cyc}(a+b)^2$ instead of $\sum\limits_{cyc}(a+b)^2$. But it is an interesting problem as it gives some sort of relationship between A.M, G.M and H.M for 3 positive numbers (we know for 2 anyway). Please see my answer below where I have represented the problem in form of A.M, G.M and H.M. – Math Lover Jul 05 '20 at 20:18
  • Two times Muirhead) – Alexey Burdin Jul 06 '20 at 01:07
  • @Math Lover It was typo. I fixed. Thank you! – Michael Rozenberg Jul 06 '20 at 04:28
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Due to homogeneity, assume that $abc = 1$.

Let $p = a+b+c, q = ab+bc+ca, r=abc=1$.

We need to prove that $p\cdot \frac{q}{r} + 3 \ge 4 \cdot \frac{p}{\sqrt[3]{r}}$ or $pq + 3 \ge 4 p$.

Since $q^2 \ge 3pr$, it suffices to prove that $p \cdot \sqrt{3p} + 3 \ge 4p$ or $\frac{1}{3}(\sqrt{3p} - 3)(3p - \sqrt{3p} - 3)\ge 0$ which is true since $p\ge 3$. We are done.

River Li
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Suppose the inequality is true,then, $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$$$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac3{a+b+c}\ge 4\cdot \frac{1}{\sqrt[3]{abc}}$$ $$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac3{a+b+c}\ge 4\cdot \frac{1}{\sqrt[3]{abc}}\geq \frac{12}{a+b+c}$$ $$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq \frac {9}{a+b+c} $$$$\Rightarrow \frac {a+b+c}{3}\geq\frac{3}{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$$Which is obviously true ($AM\geq HM $).

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    In the line where you compared with $\frac{12}{a+b+c}$ is not correct. If you can fit something in the middle, then it is allowed. – farruhota Jul 09 '20 at 06:15