Welcome to our community

Be a part of something great, join today!

The Inequality Challenge

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
Prove \(\displaystyle 2^{\frac{1}{3}}+2^{\frac{2}{3}}<3\).
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
$2 = 128/64 > 125/64$, 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 $2^{2/3} + 2^{1/3} < 3$.
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
$2 = 128/64 > 125/64$, 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 $2^{2/3} + 2^{1/3} < 3$.
Thanks for participating, Opalg! I really admire your talent in approaching this type of problem using the way you did.

My solution:

Let \(\displaystyle y=2^{\frac{1}{3}}+2^{\frac{2}{3}}\). We're then asked to proved that $y<3$.

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.
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
Thanks for participating, Opalg! I really admire your talent in approaching this type of problem using the way you did.

My solution:

Let \(\displaystyle y=2^{\frac{1}{3}}+2^{\frac{2}{3}}\). We're then asked to proved that $y<3$.

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.
anemone,
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)