I have encountered the following question in a highschool book in the subject of powers. and, it seems I can't solve it....
Determine (without using a calculator) which of the following is bigger: $1+\sqrt[3]{2}$ or $\sqrt[3]{12}$
Any ideas?
Thank you! Shir
There is a known inequality between the arithmetic mean and the cubic mean. It says that for nonnegative $x$ and $y$ we have: $$ \frac{x+y}{2} \leq \sqrt[3]{\frac{x^3+y^3}{2}}. $$ If we set $x = 1$ and $y = \sqrt[3]{2}$, we get $$ \frac{1 + \sqrt[3]{2}}{2} \leq \sqrt[3]{\frac{3}{2}}. $$ By multiplying both parts by $2$ we get $1 + \sqrt[3]{2} \leq \sqrt[3]{12}$.
Assume that $1+\root 3\of 2\geq \root 3 \of {12}$. Cubing gives $$1+3\root 3\of 2 + 3\root 3\of 4 + 2\geq 12\ ,$$ or $$\root 3\of 2+\root 3\of 4\geq 3\ .\tag{1}$$ Now from $2000<2197=13^3$ we can deduce that $\root 3\of 2<1.3$, whence $\root 3\of 4< 1.69$. The last two facts together contradict $(1)$; therefore our original assumption was false. It follows that $$1+\root 3\of 2\leq \root 3 \of {12}\ .$$
Some times us mathematicians tend to avoid thinking "easily" first =) $$\sqrt[3]{12}-\sqrt[3]{2}=\sqrt[3]{2}(\sqrt[3]{6}-1)\geq 1$$
Let $y=(1+\sqrt[3]{2})^3$, and we want to know if it is bigger than 12.
$(y-3)^3/54=y$, but $(12-3)^3/54=13{1\over 2}$
$z=(x-3)^3/54-x$ is concave up if $x>3$, $z(3)<0$ so $z(x)$ has one zero (namely $y$) for $x>3$.
$z(12)=3/2>0$ so $12>y$
After reading your helpfull answers, got something of my own: $ (1+\sqrt[3]{2})<\sqrt[3]{12} \iff (1+\sqrt[3]{2})^3<12 \iff 1+3\sqrt[3]{2}+3\sqrt[3]{4} +2 < 12 \iff 3\sqrt[3]{2}+3\sqrt[3]{4} < 9 \iff \sqrt[3]{2}+\sqrt[3]{4} < 3 \iff 2+3\sqrt[3]{2}+3\sqrt[3]{4}+4 < 27 \iff 3\sqrt[3]{2}+3\sqrt[3]{4} < 21 \iff \sqrt[3]{2} + \sqrt[3]{4} < 7 $ and the last inequality is true since $\sqrt[3]{2}<2$ and $\sqrt[3]{4}<4$
Thank you all for your answers!!