How to prove $\sqrt[3]{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}}\ge \frac{\sqrt[3]{abc}+2}{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}}.$

Question

If $a,b,c>0: a+b+c=3$ then$$\sqrt[3]{\frac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{3}}\ge \frac{\sqrt[3]{abc}+2}{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}}.$$

I can prove the following inequality$$\sqrt[3]{\frac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{3}}\ge \frac{3\sqrt[3]{abc}}{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}}.$$It is weaker than the starting one because $\sqrt[3]{abc}+2\ge 3\sqrt[3]{abc}\iff 1\ge \sqrt[3]{abc}.$

We may use Cauchy-Schwarz $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}=3$$and by AM-GM $$\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}\ge 3\sqrt[9]{a^2b^2c^2}\ge 3\sqrt[3]{abc}.$$ But I can not use same idea for the starting inequality.

Can you help me? Thank you.

Update.

I also tried AM-GM $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge \frac{3}{\sqrt[3]{abc}}$$ Thus, we need to prove $$\frac{1}{\sqrt[9]{abc}}\ge \frac{\sqrt[3]{abc}+2}{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}}.$$ But I don't know how to continue the proof.

Answer 1

Proof.

Indeed, we'll prove

$$\sqrt[3]{\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\cdot (a^3+b^3+c^3)^2}\ge \frac{3abc+2(a^3+b^3+c^3)}{ab+bc+ca}.$$Based on symmetric principle, we may prove$$\sqrt[3]{b^3c^3\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\cdot (a^3+b^3+c^3)^2}\ge b^3+c^3+abc,$$or$$\sqrt[3]{\left(\frac{b^3c^3}{a^3}+b^3+c^3\right)\cdot (a^3+b^3+c^3)\cdot (a^3+b^3+c^3)}\ge b^3+c^3+abc,$$which is true by Holder inequality \begin{align*} \sqrt[3]{\left(\frac{b^3c^3}{a^3}+b^3+c^3\right)\cdot (a^3+b^3+c^3)\cdot (a^3+b^3+c^3)}&\ge \sqrt[3]{\frac{b^3c^3}{a^3}\cdot a^3\cdot a^3} +\sqrt[3]{b^3\cdot b^3\cdot b^3}+\sqrt[3]{c^3\cdot c^3\cdot c^3}\\&=b^3+c^3+abc. \end{align*} Hence, the proof is done. Equality holds iff $a=b=c=1.$

Answer 2

Either the problem is simple or I cannot find my mistake.

Do a homogenization by $a\to \frac{1}{x^3}$ etc and it suffices to prove: $$\left(\sum x^3y^3\right)^2\left(\sum x \right)^3\left(\sum x^3\right)\geq\left(2\sum x^3y^3 + 3x^2y^2z^2\right)^3.$$ WLOG assume now $x\geq y\geq z$ and use Holder for the first and third product term in the LHS: $$\sum x^3y^3\sum x^3y^3\sum x^3\geq \left(\sum_{cyc} x^3y^2\right)^3$$ so now after taking cube roots from both sides, we need to show: $$LHS = \sum x\sum_{cyc} x^3y^2\geq 2\sum x^3y^3 + 3x^2y^2z^2 = RHS$$ But after opening the brackets and simplifying, one can easily see that this is the same as: $$LHS - RHS = \left(\sum_{cyc} x^4y^2-\sum x^3y^3\right) + xyz\left(\sum_{cyc} x^2y-3xyz\right)\geq 0.$$ Now we are basically done since both summands above are non-negative. The second one is just AM-GM and the first one follows from our assumption of $x\geq y\geq z:$ $$2\left(\sum_{cyc} x^4y^2-\sum x^3y^3\right)\geq \sum_{cyc} x^4y^2 + \sum_{cyc} x^2y^4 - 2\sum x^3y^3 = \sum x^2y^2(x-y)^2\geq 0.$$

Answer 3

We have $$ \begin{aligned} &\sqrt[3]{ab}\sqrt[3]{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}}=\sqrt[3]{\frac{a+b+\frac{ab}{c}}{3}}\\ ={}&\frac1{3}\sqrt[3]{(a+b+c)(a+b+c)\Big(a+b+\frac{ab}{c}\Big)}\geqslant\frac1{3}(a+b+\sqrt[3]{abc}). \end{aligned} $$ Hence $$ \begin{aligned} &(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac})\sqrt[3]{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}}\geqslant\frac23(a+b+c)+\sqrt[3]{abc}\\ \iff{}& \begin{aligned} \sqrt[3]{\frac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{3}}\ge \frac{\sqrt[3]{abc}+2}{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}} \end{aligned}. \end{aligned} $$