Is it inequality true

Question

Please help me prove the inequality:

$$\sqrt[3]{3+\sqrt[3] 3}+\sqrt[3]{3-\sqrt[3] 3}<2\sqrt[3] 3$$

Thanky for your help and your attention.

Answer 1

Let be $$a=\sqrt[3]{3+\sqrt[3] 3}, b=\sqrt[3]{3-\sqrt[3] 3}$$ now we have $$a^3+b^3=\left(\sqrt[3]{3+\sqrt[3] 3}\right)^3+\left(\sqrt[3]{3-\sqrt[3] 3}\right)^3=3+\sqrt[3]3+3-\sqrt[3]3=6.$$ Becouese, $$a^3+b^3>a^2b+b^2a$$ we have $$ab(a+b)<a^3+b^3=6$$ Now $$3ab(a+b)<18$$ or $$3a^2b+3ab^2<18$$ Add in the both sides eqution: $$a^3+b^3=6$$ and inequality $$3a^2b+3ab^2<18$$ we have: $$(a+b)^3<24=8\cdot 3$$ or $$a+b<2\sqrt[3]3$$

Answer 2

We can use $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$.

Since $a^2+b^2+c^2-ab-ac-bc=0\Leftrightarrow a=b=c$

and $\sqrt[3]{3+\sqrt[3]3}\neq\sqrt[3]{3-\sqrt[3]3}$ , we need to prove that $$3+\sqrt[3]3+3-\sqrt[3]3-8\cdot3+3\cdot2\sqrt[3]3\sqrt[3]{9-\sqrt[3]9}<0$$ or $$\sqrt[3]{27-3\sqrt[3]9}<3$$ which is obvious.

Answer 3

Also your inequality it's just Jensen for $f(x)=\sqrt[3]x$.

Indeed, $f$ is a concave function.

Since your inequality it's $$\frac{\sqrt[3]{3+\sqrt[3] 3}+\sqrt[3]{3-\sqrt[3] 3}}{2}<\sqrt[3]{\frac{3+\sqrt[3] 3+3-\sqrt[3] 3}{2}}$$ we are done!