Show that $\displaystyle \sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}$
I was thinking along the lines of dividing both sides by $\displaystyle \sqrt[3]{3}$, rearranging and comparing the cubes of both sides but doesnt seem to work.
Show that $\displaystyle \sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}$
I was thinking along the lines of dividing both sides by $\displaystyle \sqrt[3]{3}$, rearranging and comparing the cubes of both sides but doesnt seem to work.
There is probably a simpler proof, but here is what i suggest:
We go to a proof by contradiction. Assume that is valid:
$\displaystyle
\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}} > 2\sqrt[3]{3}$
We raise both sides to cube:
$\displaystyle
(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}})^3 > (2\sqrt[3]{3})^3$
By binomial expansion we have:
$\displaystyle
6+3(3-3^{1/3})^{2/3}(3+3^{1/3})^{1/3}+3(3-3^{1/3})^{1/3}(3+3^{1/3})^{2/3} > 24$
By extracting as common factor the term:
$\displaystyle (3^2-3^{2/3})^{1/3}$ we do have:
$\displaystyle
(3^2-3^{2/3})^{1/3}(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}})> 18$
But we know that $\displaystyle \sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}} > 2\sqrt[3]{3}$
So by substituting we have:
$\displaystyle
(3^2-3^{2/3})^{1/3}2\sqrt[3]{3} > 18
$
Or
$\displaystyle
6(3^3-3\sqrt[3]{3})^{1/3} > 18\Rightarrow (3^3-3\sqrt[3]{3})^{1/3} > 3
$
By raising again to cube we have that:
$\displaystyle
3^3-3\sqrt[3]{3} > 3^3\Rightarrow-3\sqrt[3]{3} > 0
$
which is false.
If you know what that means, you can use the strict concavity of the cubic root function. Id est: $\displaystyle \sqrt[3]{\frac{x+y}{2}}>\frac{\sqrt[3]{x}+\sqrt[3]{y}}{2}$ when $\displaystyle 0\leq x\neq y$. With $\displaystyle x,y=3\pm\sqrt[3]{3}$. This is the "visual" proof (you can see the inequality on a plot of $\displaystyle \sqrt[3]{}$) and it immediately generalizes to $\displaystyle \sqrt[3]{a+b}+\sqrt[3]{a-b}< 2\sqrt[3]{a}$ for any $\displaystyle 0<b<a$.