MHB Can You Prove $a > \sqrt[9]{8}$ is a Root of a Polynomial with $1 < a < 2$?

[anemone]

Let $1\lt a \lt 2$, $a$ is a root of the equation $x^5-x-2=0$. Prove that $\large a>\sqrt[9]{8}$.

 

[kaliprasad]

Let $1\lt a \lt 2$, $a$ is a root of the equation $x^5-x-2=0$. Prove that $\large a>\sqrt[9]{8}$.

Spoiler

let $f(x) = x^5-x-2=0$ then we have $f(1) = - 2$ , $f(2)= 28$ and $\frac{df}{dx} = 5x^4 -1 >=0$ for $ x> 1$
so f(1) is -ve and f(2) is + ve and f(x) is increasing in given range from 1 2 (other value
$>$ 1 also but we are not bothered)
if $a > \sqrt[9]8$ for a to be root $f(\sqrt[9]8 = \sqrt[3]2)$ is -ve
$f(\sqrt[3]2) = \sqrt[3]2^5 - \sqrt[3]2 -2= 2 *\sqrt[3]2^2 - \sqrt[3]2 - 2\cdots(1)$
Now $\frac{1}{2}(2 + \sqrt[3]2) >= \sqrt{2 \sqrt[3]2}= \sqrt[3]2^2$
or $2 + \sqrt[3]2 >= 2\sqrt[3]2^2$
from (1) and above we have $f(\sqrt[3]2) < 0 $ hence $a > \sqrt[9]8$

 

[anemone]

Well done, kaliprasad! And thanks for participating!(Cool)

My solution:

Spoiler

Since $a$ is a root of the equation $x^5-x-2=0$, we get the relation:

$a^5-a-2=0$

$a^5=a+2$(*)

Since $a$ is a positive real, we can apply the AM-GM inequality to the expression:

$a+2\ge 2\sqrt{2a}$

Now, we replace (*) into the inequality above to obtain:

$a^5\ge 2\sqrt{2a}$

Squaring both sides yields:

$a^{10}\ge 8a$

Again, since $a$ is a positive real, we can divide through the inequality above by $a$, and then taking ninth root on both sides of the inequality and the result follows:

$a^9\ge 8$

$a\ge \sqrt[9]{8}$ (Q.E.D.)

 

[kaliprasad]

Well done, kaliprasad! And thanks for participating!(Cool)

My solution:

Spoiler

Since $a$ is a root of the equation $x^5-x-2=0$, we get the relation: $a\ge \sqrt[9]{8}$ (Q.E.D.)

Spoiler

we are supposed to prove $a\ > \sqrt[9]{8}$

 

[anemone]

Spoiler

we are supposed to prove $a\ > \sqrt[9]{8}$

Ops...it was my bad...I should have realized the strict inequality should be in place but not including the "equal to" sign...

I will re-post my solution here of the correct version of my solution:

My solution:

Spoiler

Since $a$ is a root of the equation $x^5-x-2=0$, we get the relation:

$a^5-a-2=0$

$a^5=a+2$(*)

Since $a$ is a positive real, we can apply the AM-GM inequality to the expression:

$a+2\gt 2\sqrt{2a}$ (since $a\ne 2$)

Now, we replace (*) into the inequality above to obtain:

$a^5\gt 2\sqrt{2a}$

Squaring both sides yields:

$a^{10}\gt 8a$

Again, since $a$ is a positive real, we can divide through the inequality above by $a$, and then taking ninth root on both sides of the inequality and the result follows:

$a^9\gt 8$

$a\gt \sqrt[9]{8}$ (Q.E.D.)