Prove this inequality $\sqrt[3]{(a+b+c)(b+c+d)}\ge\sqrt[3]{ab}+\sqrt[3]{cd}$

Question

Let $a,b,c,d>0$ show that $$\sqrt[3]{(a+b+c)(b+c+d)}\ge\sqrt[3]{ab}+\sqrt[3]{cd}$$

Idea: Hence, we need to prove that $$(a+b+c)(b+c+d)\ge ab+cd+3\sqrt[3]{(ab)^2cd}+3\sqrt[3]{(cd)^2(ab)}$$ $$\Longleftrightarrow ac+bc+bd+ad+b^2+c^2\ge 3\sqrt[3]{(ab)^2cd}+3\sqrt[3]{(cd)^2(ab)}$$ I attempted a following proof by AM-GM inequality,But I am not able to solve the upper part, this inequality

Answer 1

It's just Holder: $$(a+b+c)(b+c+d)\left(\frac{b}{b+c}+\frac{c}{b+c}\right)\geq$$ $$\geq\left(\sqrt[3]{a\cdot(b+c)\cdot\frac{b}{b+c}}+\sqrt[3]{(b+c)\cdot d\cdot\frac{c}{b+c}}\right)^3=\left(\sqrt[3]{ab}+\sqrt[3]{cd}\right)^3$$