Prove that: $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+5\ge (a+b)(b+c)(c+a)$

Question

Let $a,b,c>0$ satisfy $a+b+c=3$ Prove that: $$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+5\ge (a+b)(b+c)(c+a)$$I have my solution, we need to prove: $$3\sqrt[3]{a}+3\sqrt[3]{b}+3\sqrt[3]{c}+15\ge 3(a+b)(b+c)(c+a)$$ We know that: $$3(a+b)(b+c)(c+a)=(a+b+c)^3-a^3-b^3-c^3$$ So the problem is: $$a^3+3\sqrt[3]{a}+b^3+3\sqrt[3]{b}+c^3+3\sqrt[3]{c}+15\ge 27$$ By AM-GM, we have: $$a^3+3\sqrt[3]{a}+b^3+3\sqrt[3]{b}+c^3+3\sqrt[3]{c}\ge4a+4b+4c=12$$ and we done.

Is there any other way? please help me

Answer 1

Another way.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, $v^2\leq u^2=1$ and by Schur $w^3\geq4uv^2-3u^3=4v^2-3$.

Id est, by AM-GM and C-S we obtain: $$5-\prod_{cyc}(a+b)+\sum_{cyc}\sqrt[3]a=5-(9uv^2-w^3)+\sum_{cyc}\frac{3a}{3\sqrt[3]{a^2}}\geq$$ $$\geq5-9v^2+4v^2-3+3\sum_{cyc}\frac{a}{2a+1}=2-5v^2+3\sum_{cyc}\frac{a^2}{2a^2+a}\geq$$ $$\geq2-5v^2+\frac{3(a+b+c)^2}{\sum\limits_{cyc}(2a^2+a)}=2-5v^2+\frac{27}{2(9u^2-6v^2)+3}=$$ $$=2-5v^2+\frac{9}{7-4v^2}=\frac{(1-v^2)(23-20v^2)}{7-4v^2}\geq0.$$ The following stronger inequality is also true.

Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that: $$\sqrt[3]a+\sqrt[3]b+\sqrt[3]c+2\geq\frac{5}{8}(a+b)(a+c)(b+c).$$