MHB Cube Root Challenge: Prove Inequality

AI Thread Summary
The discussion centers on proving the inequality $\sqrt[3]{43}<\sqrt[3]{9}+\sqrt[3]{3}<\sqrt[3]{44}$. Participants agree that the proof is straightforward, with one user mentioning the use of the arithmetic mean-geometric mean inequality (AM-GM) to establish that $x>\dfrac{31}{9}$. The conversation emphasizes the necessity of the proof, indicating a shared understanding of its importance. Overall, the participants are engaged in mathematical reasoning to validate the inequality. The discussion highlights both the simplicity and significance of the proof in the context of cube roots.
anemone
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Prove $\sqrt[3]{43}<\sqrt[3]{9}+\sqrt[3]{3}<\sqrt[3]{44}$
 
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anemone said:
Prove $\sqrt[3]{43}<\sqrt[3]{9}+\sqrt[3]{3}<\sqrt[3]{44}----(1)$
my solution :
if (1) is true then cube and rearrange it we get :
$31<9(\sqrt[3]{9}+\sqrt[3]{3})<32$
or $31<9(x^2+x)<32$
or $31<9x(x+1)<32---(2)$
here we let $x=\sqrt[3]{3}>1$
if $1.44<x=\sqrt[3]{3}<1.45 $ then both sides of (2)will be satisfied
checking wih calculator $\sqrt[3]{3}\approx 1.44225$
in fact we can apply $AP>GP$ to both sides of (2) and get the same result
 
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Thanks Albert for participating...

I think you have to explicitly prove that $1.44<x=\sqrt[3]{3}<1.45 $ is true to complete your proof...:)
 
anemone said:
Thanks Albert for participating...

I think you have to explicitly prove that $1.44<x=\sqrt[3]{3}<1.45 $ is true to complete your proof...:)
it is easy :
$1.44^3=2.985984<3<1.45^3=3.048625$
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Albert said:
it is easy :
$1.44^3=2.985984<3<1.45^3=3.048625$

Yes, it's easy and it's necessary. :D

My solution:

Let $x= \sqrt[3]{9} +\sqrt[3]{3}$, cube the inequality and rearrange, we have to prove $\dfrac{31}{9}\lt x \lt \dfrac{32}{9}$, but note that

$9(20^3)=72000\gt 41^3=68921$, this gives $\sqrt[3]{9} \gt \dfrac{41}{20}$ and

$3(25^3)=46875\gt 36^3=46656$, this gives $\sqrt[3]{3} \gt \dfrac{36}{25}$, so $x>\dfrac{41}{20}+\dfrac{36}{25}=3\dfrac{49}{100}>\dfrac{31}{9}$

Also,

$9(10^3)=9000< 21^3=9261$, this gives $\sqrt[3]{9} < \dfrac{21}{10}$ and

$3(20^3)=24000< 29^3=24389$, this gives $\sqrt[3]{3}<\dfrac{29}{20}$, so $x<\dfrac{21}{20}+\dfrac{29}{20}=2\dfrac{1}{2}<\dfrac{32}{9}$

so we're done.
 
anemone said:
Yes, it's easy and it's necessary. :D

My solution:

Let $x= \sqrt[3]{9} +\sqrt[3]{3}$, cube the inequality and rearrange, we have to prove $\dfrac{31}{9}\lt x \lt \dfrac{32}{9}$, but note that

$9(20^3)=72000\gt 41^3=68921$, this gives $\sqrt[3]{9} \gt \dfrac{41}{20}$ and

$3(25^3)=46875\gt 36^3=46656$, this gives $\sqrt[3]{3} \gt \dfrac{36}{25}$, so $x>\dfrac{41}{20}+\dfrac{36}{25}=3\dfrac{49}{100}>\dfrac{31}{9}$

Also,

$9(10^3)=9000< 21^3=9261$, this gives $\sqrt[3]{9} < \dfrac{21}{10}$ and

$3(20^3)=24000< 29^3=24389$, this gives $\sqrt[3]{3}<\dfrac{29}{20}$, so $x<\dfrac{21}{20}+\dfrac{29}{20}=2\dfrac{1}{2}<\dfrac{32}{9}$

so we're done.
I will use $AP>GP$ to prove $x>\dfrac{31}{9}$
let :$x= \sqrt[3]{9} +\sqrt[3]{3}$
$x>2\sqrt 3>\dfrac {31}{9}$
for $\sqrt 3>1.73>\dfrac{31}{18}$
 
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