Let a, b > 0, a+b = 1, n>1 Show that (a+1/a)^n + (b+1/b)^n >= 5^n/n^(n-1)?

Let $a , b > 0 , a + b = 1 , n > 1$ Show that ${\left(a + \frac{1}{a}\right)}^{n} + {\left(b + \frac{1}{b}\right)}^{n} \ge {5}^{n} / {n}^{n - 1}$

Mar 12, 2017

See below.

Explanation:

This problem can be stated as a minimization problem.

Calling $f \left(a , b\right) = {\left(a + \frac{1}{a}\right)}^{n} + {\left(b + \frac{1}{b}\right)}^{n}$

Find $\left\{{a}_{0} , {b}_{0}\right\} = \text{arg} \min f \left(a , b\right)$

subjected to

$g \left(a , b\right) = a + b = 1$

If ${\left({a}_{0} + \frac{1}{a} _ 0\right)}^{n} + {\left({b}_{0} + \frac{1}{b} _ 0\right)}^{n} \ge {5}^{n} / {n}^{n - 1}$ then the proposition will be true.

Analyzing the problem we see due to the symmetry, that $a , b$ will require the same resources so a good guess is ${a}_{0} = \frac{1}{2} , {b}_{0} = \frac{1}{2}$

Now substituting those values into the objective function we have

${\left({a}_{0} + \frac{1}{a} _ 0\right)}^{n} + {\left({b}_{0} + \frac{1}{b} _ 0\right)}^{n} = {5}^{n} / {2}^{n - 1} \ge {5}^{n} / {n}^{n - 1}$ for $n > 1$

It is necessary to analyze for the minimum of $f \left(a , b\right)$ at ${a}_{0} , {b}_{0}$

The Hessian ${H}_{f} = \left(\begin{matrix}{f}_{a & a} & {f}_{a & b} \\ {f}_{b & a} & {f}_{b & b}\end{matrix}\right)$

at ${a}_{0} , {b}_{0}$ is

${H}_{f} = \left(\begin{matrix}{\left(\frac{2}{5}\right)}^{2 - n} n \left(31 + 9 n\right) & 0 \\ 0 & {\left(\frac{2}{5}\right)}^{2 - n} n \left(31 + 9 n\right)\end{matrix}\right)$

This matrix has as characteristic polynomial

$p \left(\lambda\right) = {\left(\lambda - {\left(\frac{2}{5}\right)}^{2 - n} n \left(31 + 9 n\right)\right)}^{2}$ with positive roots so ${a}_{0} , {b}_{0}$ is a minimum point, and the assertion is true.

Note. The point ${a}_{0} , {b}_{0}$ is a stationary point because with

$\frac{d}{\mathrm{da}} \left(f \circ g\right) \left(a\right) = \left(\frac{1}{a - 1} ^ 2 - 1\right) {\left(1 + \frac{1}{1 - a} - a\right)}^{n - 1} n + \left(1 - \frac{1}{a} ^ 2\right) {\left(\frac{1}{a} + a\right)}^{n - 1} n$

we have $\frac{d}{\mathrm{da}} \left(f \circ g\right) \left(\frac{1}{2}\right) = 0$

Mar 14, 2017

See below.

Explanation:

Another point of view.

Making now $u = a + \frac{1}{a}$ and $v = b + \frac{1}{b}$ we have

${u}^{n} + {v}^{n} \ge {5}^{n} / \left({n}^{n - 1}\right)$ subjected to

$\frac{1}{2} \left(u \pm \sqrt{{u}^{2} - 4}\right) + \frac{1}{2} \left(v \pm \sqrt{{v}^{2} - 4}\right) = 1$

Due to the symmetry the minimization problem requires that

$\frac{1}{2} \left(u \pm \sqrt{{u}^{2} - 4}\right) = \frac{1}{2} \left(v \pm \sqrt{{v}^{2} - 4}\right) = \frac{1}{2}$

or

${\left(u - 1\right)}^{2} = {u}^{2} - 4 \to u = {u}_{0} = \frac{5}{2}$
${\left(v - 1\right)}^{2} = {v}^{2} - 4 \to v = {v}_{0} = \frac{5}{2}$

and

${u}_{0}^{n} + {v}_{0}^{n} = 2 {\left(\frac{5}{2}\right)}^{n} = {5}^{n} / \left({2}^{n - 1}\right) \ge {5}^{n} / \left({n}^{n - 1}\right) \forall n > 1$

Suppose instead that symmetry does not occur

$\frac{1}{2} \left({u}_{0} \pm \sqrt{{u}_{0}^{2} - 4}\right) = \frac{1}{2} + \epsilon$ and $\frac{1}{2} \left({v}_{0} \pm \sqrt{{v}_{0}^{2} - 4}\right) = \frac{1}{2} - \epsilon$

then

$\left\{\begin{matrix}{u}_{0} = \frac{4 + {\left(1 + 2 \epsilon\right)}^{2}}{2 \left(1 + 2 \epsilon\right)} \\ {v}_{0} = \frac{4 + {\left(1 - 2 \epsilon\right)}^{2}}{2 \left(1 - 2 \epsilon\right)}\end{matrix}\right.$

and

$f \left(\epsilon\right) = {\left(\frac{4 + {\left(1 + 2 \epsilon\right)}^{2}}{2 \left(1 + 2 \epsilon\right)}\right)}^{n} + {\left(\frac{4 + {\left(1 - 2 \epsilon\right)}^{2}}{2 \left(1 - 2 \epsilon\right)}\right)}^{n}$

Now

(df)/(d epsilon)=((4 epsilon (1 + epsilon)-3) (1/2 + epsilon + 2/(1 + 2 epsilon))^(n-1) n)/(1 + 2 epsilon)^2-((1/2 + 2/(1 - 2 epsilon) - epsilon)^(n-1) (2 epsilon-3) (1 + 2 epsilon) n)/(1 - 2 epsilon)^2

and as can be verified,

$\frac{\mathrm{df}}{d \epsilon} \left(0\right) = 0$ and also

$\frac{{d}^{2} f}{d {\epsilon}^{2}} \left(0\right) = {5}^{n - 2} / {2}^{n - 3} n \left(31 + 9 n\right) > 0$ so $\epsilon = 0$ is a minimum and this result confirms ${u}_{0} = {v}_{0} = \frac{5}{2}$ as the solution point, validating the proposition.