Proving $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ for positive $a$, $b$, $c$

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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 $).