The inequality challenge of proving that $$2^{\frac{1}{3}} + 2^{\frac{2}{3}} < 3$$ has been effectively demonstrated through a series of mathematical deductions. The proof utilizes the properties of cube roots and inequalities, establishing that $$2^{1/3} > 5/4$$ and consequently $$\dfrac{1}{2^{1/3} - 1} < 4$$. This leads to the conclusion that $$2^{2/3} + 2^{1/3} < 3$$, confirming the original inequality. The discussion also highlights the need for further exploration of potential solutions to the cubic equation derived from the inequality.
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Opalg said:[sp]$2 = 128/64 > 125/64$, so (taking cube roots) $2^{1/3} > 5/4$ and $2^{1/3} -1 >1/4$. Therefore $\dfrac1{2^{1/3} -1} < 4$. But $$1 = 2-1 = (2^{1/3})^3 - 1 = (2^{1/3} -1)(2^{2/3} + 2^{1/3} + 1),$$ and so $2^{2/3} + 2^{1/3} + 1 = \dfrac1{2^{1/3} -1} < 4$. Thus $2^{2/3} + 2^{1/3} < 3$.[/sp]
anemone said:Thanks for participating, Opalg! I really admire your talent in approaching this type of problem using the way you did.
My solution:
Let $$y=2^{\frac{1}{3}}+2^{\frac{2}{3}}$$. We're then asked to proved that $y<3$.
Then $$y^3=2+3(2^{\frac{1}{3}})(2^{\frac{2}{3}})(2^{\frac{1}{3}}+2^{\frac{2}{3}})+2^2=6+6y$$
$$y^3-6y-6=0$$
If we let $f(y)=y^3-6y-6$, we see that $f(2)=-10$ and $f(3)=3$, hence by the Intermediate Value Theorem, $y$ must have a solution between 2 and 3, i.e. $y<3$ and so we're done.