Cube Root Challenge: Prove Inequality

[anemone]

Prove $\sqrt[3]{43}<\sqrt[3]{9}+\sqrt[3]{3}<\sqrt[3]{44}$

 

[Albert1]

Prove $\sqrt[3]{43}<\sqrt[3]{9}+\sqrt[3]{3}<\sqrt[3]{44}----(1)$

my solution :

Spoiler

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

 

[anemone]

Thanks Albert for participating...

Spoiler

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

 

[Albert1]

Thanks Albert for participating...

Spoiler

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 :

Spoiler

$1.44^3=2.985984<3<1.45^3=3.048625$

 

[anemone]

it is easy :

Spoiler

$1.44^3=2.985984<3<1.45^3=3.048625$

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

My solution:

Spoiler

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.

 

[Albert1]

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

My solution:

Spoiler

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 $x>\dfrac{31}{9}$

Spoiler

let :$x= \sqrt[3]{9} +\sqrt[3]{3}$
$x>2\sqrt 3>\dfrac {31}{9}$
for $\sqrt 3>1.73>\dfrac{31}{18}$