Is it inequality true 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.

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
$$a+b<2\sqrt[3]3$$
