algebra :)

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by hemangsarkar, Sep 1, 2012, 7:35 PM

Q) let $a$ be a positive real number such that $a^5 - a^3 + a = 2$ ,
show that $3 < a^6 < 4$ .

my solution :

$(a^6 + 1) = (a^2+1)(a^4-a^2+1) = \frac{a^2+1}{a}*(a^5-a^3+a) = 2\left(a + \frac{1}{a} \right)$

but $a + \frac{1}{a} \geq 2$

the last inequality holds by am - gm.

hence $a^6 + 1 \geq 4$ or, $a^6 \geq 3$ .

note that equality doesn't hold since $a = 1$ isn't a solution to

$a^5 - a^3 + a = 2$ .

to prove that $a^6 < 4$ .

we have $a^3 + 2 = a^5 + a$

or, $1 + \frac{2}{a^3} = a^2 + \frac{1}{a^2} \geq 2$ .

again equality doesn't hold.

so, we must have $a^3 < 2$ , or $a^6 < 4$ .

done.

This post has been edited 3 times. Last edited by hemangsarkar, Sep 5, 2012, 3:19 PM

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algebra :)

by hemangsarkar, Sep 1, 2012, 7:35 PM

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