Can the Inequality Challenge be Proven: 2^{\frac{1}{3}}+2^{\frac{2}{3}}<3?

[anemone]

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

 

[Opalg]

[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]

[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]

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

My solution:

Spoiler

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]

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

My solution:

Spoiler

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)

 

[CellCoree]

I would approach this statement by first breaking down the equation and evaluating each component individually.

Starting with 2^{\frac{1}{3}}, we can rewrite this as the cube root of 2. Using a calculator, we can see that the cube root of 2 is approximately 1.2599.

Next, we look at 2^{\frac{2}{3}}. This can be rewritten as the cube root of 2 squared, which is equal to the cube root of 4. Using a calculator, we can see that the cube root of 4 is approximately 1.5874.

Now, we can substitute these values back into the original equation: 1.2599 + 1.5874. Using basic arithmetic, we can see that this is equal to 2.8473.

Lastly, we need to compare this value to 3. We can see that 2.8473 is less than 3, therefore proving that 2^{\frac{1}{3}}+2^{\frac{2}{3}}<3.

In conclusion, as a scientist, I can confidently say that the statement is true and can be proven mathematically. This inequality challenge highlights the importance of understanding and evaluating mathematical equations, as well as the significance of precision and accuracy in scientific research.