I have proved the following inequality:
Let $a,b,c>0$ $$\dfrac{(a+\sqrt{ab}+\sqrt[3]{abc})}{3}\le \sqrt[3]{a\cdot\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3}}$$ My solution is:$$a\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3} =\dfrac{1}{3^3}(a+a+a)(a+\dfrac{a+b}{2}+b)(a+b+c)\ge\dfrac{1}{3^3}(a+a+a)(a+\sqrt{ab}+b)(a+b+c)\ge \dfrac{1}{3}(a+\sqrt{ab}+\sqrt[3]{abc})^3$$
But I met the following, harder inequality: $$3(a+\sqrt{ab}+\sqrt[3]{abc})\le \left(7+\dfrac{4\sqrt{ab}}{a+b}\right)\sqrt[3]{a\cdot\dfrac{a+b}{2}\cdot\dfrac{a+b+c}{3}}$$ How to prove this inequality? Thanks