# Prove that if 1<a<2 , that 2<a+1/a<3? please , i really need your help!!!

Apr 16, 2018

See explanation

#### Explanation:

Let $a = \frac{p}{q}$ where $p$ and $q$ are positive integers.

$1 < \frac{p}{q}$ therefore $q < p$. $\frac{p}{q} < 2$ therefore $p < 2 q$. Therefore $q < p < 2 q$.

$a + \frac{1}{a} = \frac{p}{q} + \frac{q}{p} = \frac{p p}{q p} + \frac{q q}{p q} = \frac{{p}^{2} + {q}^{2}}{p q} = \frac{{p}^{2} + 2 p q + {q}^{2} - 2 p q}{p q} = {\left(p + q\right)}^{2} / \left(p q\right) - \frac{2 p q}{p q} = {\left(p + q\right)}^{2} / \left(p q\right) - 2$

${\left(q + q\right)}^{2} / \left(q q\right) < {\left(p + q\right)}^{2} / \left(p q\right) < {\left(2 q + q\right)}^{2} / \left(2 q q\right)$*
${\left(2 q\right)}^{2} / {q}^{2} < {\left(p + q\right)}^{2} / \left(p q\right) < {\left(3 q\right)}^{2} / \left(2 {q}^{2}\right)$
$\frac{4 {q}^{2}}{q} ^ 2 < {\left(p + q\right)}^{2} / \left(p q\right) < \frac{9 {q}^{2}}{2 {q}^{2}}$
$4 < {\left(p + q\right)}^{2} / \left(p q\right) < \frac{9}{2}$
$4 - 2 < {\left(p + q\right)}^{2} / \left(p q\right) - 2 < \frac{9}{2} - 2$
$2 < {\left(p + q\right)}^{2} / \left(p q\right) - 2 < \frac{5}{2}$
$2 < a + \frac{1}{a} < \frac{5}{2}$

$\frac{5}{2} < \frac{6}{2}$
$\frac{5}{2} < 3$
$2 < a + \frac{1}{a} < 3$

*This assumes that as $p$ increases, ${\left(p + q\right)}^{2} / \left(p q\right)$ increases. This can be verified intuitively, by looking at the graph of $y = {\left(x + q\right)}^{2} / \left(x q\right)$ on $x \in \left(q , 2 q\right)$ for various positive values of $q$, or by the calculus process below.

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$\frac{\partial}{\partial p} \left[{\left(p + q\right)}^{2} / \left(p q\right)\right] = \frac{1}{q} \frac{\partial}{\partial p} \left[{\left(p + q\right)}^{2} / p\right] = \frac{1}{q} \frac{p \frac{\partial}{\partial p} \left[{\left(p + q\right)}^{2}\right] - {\left(p + q\right)}^{2} \frac{\partial}{\partial p} \left[p\right]}{p} ^ 2 = \frac{1}{q} \frac{p \left[2 \left(p + q\right)\right] - {\left(p + q\right)}^{2} \left[1\right]}{p} ^ 2 = \frac{1}{q} \frac{2 p \left(p + q\right) - {\left(p + q\right)}^{2}}{p} ^ 2 = \frac{\left(2 {p}^{2} + 2 p q\right) - \left({p}^{2} + 2 p q + {q}^{2}\right)}{{p}^{2} q} = \frac{{p}^{2} - {q}^{2}}{{p}^{2} q}$.

On $p \in \left(q , 2 q\right)$:
Since $p > q > 0$, ${p}^{2} > {q}^{2}$ thus ${p}^{2} - {q}^{2} > 0$.
Since $q > 0$, ${p}^{2} q > 0$
Since ${p}^{2} - {q}^{2} > 0$ and ${p}^{2} q > 0$, $\frac{{p}^{2} - {q}^{2}}{{p}^{2} q} > 0$
Since $\frac{\partial}{\partial p} \left[{\left(p + q\right)}^{2} / \left(p q\right)\right] = \frac{{p}^{2} - {q}^{2}}{{p}^{2} q}$ and $\frac{{p}^{2} - {q}^{2}}{{p}^{2} q} > 0$, $\frac{\partial}{\partial p} \left[{\left(p + q\right)}^{2} / \left(p q\right)\right] > 0$

Therefore ${\left(p + q\right)}^{2} / \left(p q\right)$ is increasing for constant $q$ and $q < p < 2 q$ because $\frac{\partial}{\partial p} \left[{\left(p + q\right)}^{2} / \left(p q\right)\right]$ is positive.

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Apr 16, 2018

In description

#### Explanation:

Here constraint (1):

$1 < a < 2$

Constraint(2):
By reciprocal theorem,

$\frac{1}{1} > \frac{1}{a} > \frac{1}{2}$
$1 > a > \frac{1}{2}$

In constraint 1 add 1 on both sides,
$1 + 1 < a + 1 < 2 + 1$
$2 < a + 1 < 3$

$\textcolor{red}{a + 1 < 3}$

$\left(1 + \frac{1}{2}\right) < \left(a + \frac{1}{2}\right) < \left(2 + \frac{1}{2}\right)$

Again note that, $2 < 2 + \frac{1}{2}$
So $a + \frac{1}{2}$ must be less than 2
$\textcolor{red}{a + \frac{1}{2}} < 2$

Hence In constraint 2,
$1 > a > \frac{1}{2}$

$1 + a > a + \frac{1}{a} > \frac{1}{2} + a$
$3 > a + \frac{1}{a} > 2$
$2 < a + \frac{1}{a} < 3$

We did it so because $a + 1 < 3$
So $a + \frac{1}{a}$ must be less than 3.

Again $a + \frac{1}{2} < 2$ but in this constraint $a + \frac{1}{a} > a + \frac{1}{2}$
So, $a + \frac{1}{a}$ must be greater than 2.
Hence, $1 > \frac{1}{a} > 1 \setminus 2$

By adding a on both sides,
$1 + a > a + \frac{1}{a} > a + \frac{1}{2}$
$3 > a + \frac{1}{a} > 2$
$2 < a + \frac{1}{a} < 3$ proved