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Prove \(\displaystyle 2^{\frac{1}{3}}+2^{\frac{2}{3}}<3\).

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Prove \(\displaystyle 2^{\frac{1}{3}}+2^{\frac{2}{3}}<3\).

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Thanks for participating,

Then \(\displaystyle y^3=2+3(2^{\frac{1}{3}})(2^{\frac{2}{3}})(2^{\frac{1}{3}}+2^{\frac{2}{3}})+2^2=6+6y\)

\(\displaystyle y^3-6y-6=0\)

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.

- Mar 31, 2013

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anemone,Thanks for participating,Opalg!I really admire your talent in approaching this type of problem using the way you did.My solution:

Then \(\displaystyle y^3=2+3(2^{\frac{1}{3}})(2^{\frac{2}{3}})(2^{\frac{1}{3}}+2^{\frac{2}{3}})+2^2=6+6y\)

\(\displaystyle y^3-6y-6=0\)

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.

the proof provided by you is far from complete

(because if we take

f(x) = (y-1.5)(y-2.5)(y+.5) = 0

we get f(2.0) < 0 and f(3) > 0 and it has 3 roots)

based on this we need to prove in your case

either it has no other real solution or other 2 solutions if real are no where near 2^(1/3) + 2^(2/3)