Question

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 `(a+1/a)^n + (b+1/b)^n >= 5^n/n^(n-1)`

Answer 1

See below.

Explanation:

This problem can be stated as a minimization problem.

Calling `f(a,b)=(a+1/a)^n + (b+1/b)^n`

Find `{a_0,b_0}="arg"minf(a,b)`

subjected to

`g(a,b)=a+b=1`

If `(a_0+1/a_0)^n + (b_0+1/b_0)^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=1/2, b_0=1/2`

Now substituting those values into the objective function we have

`(a_0+1/a_0)^n + (b_0+1/b_0)^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(a,b)` at `a_0,b_0`

The Hessian `H_f=((f_(a,a),f_(a,b)),(f_(b,a),f_(b,b)))`

at `a_0,b_0` is

`H_f=(((2/5)^(2 - n) n (31 + 9 n), 0),(0, (2/5)^(2 - n) n (31 + 9 n)))`

This matrix has as characteristic polynomial

`p(lambda)=(lambda-(2/5)^(2 - n) n (31 + 9 n))^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

`d/(da)(f @ g)(a)=(1/(a-1)^2-1) (1 + 1/(1 - a) - a)^(n-1) n + (1 - 1/a^2) (1/a + a)^(n-1) n`

we have `d/(da)(f @ g)(1/2)=0`

Answer 2

See below.

Explanation:

Another point of view.

Making now `u=a+1/a` and `v=b+1/b` we have

`u^n+v^n ge 5^n/(n^(n-1))` subjected to

`1/2(upmsqrt(u^2-4))+1/2(vpm sqrt(v^2-4))=1`

Due to the symmetry the minimization problem requires that

`1/2(upmsqrt(u^2-4))=1/2(vpm sqrt(v^2-4))=1/2`

or

`(u-1)^2=u^2-4->u=u_0=5/2`
`(v-1)^2=v^2-4->v=v_0=5/2`

and

`u_0^n+v_0^n=2(5/2)^n= 5^n/(2^(n-1)) ge 5^n/(n^(n-1)) forall n > 1`

Suppose instead that symmetry does not occur

`1/2(u_0pmsqrt(u_0^2-4))=1/2+epsilon` and `1/2(v_0pm sqrt(v_0^2-4))=1/2-epsilon`

then

`{(u_0=(4+(1+2epsilon)^2)/(2(1+2epsilon))),(v_0=(4+(1-2epsilon)^2)/(2(1-2epsilon))):}`

and

`f(epsilon)=((4+(1+2epsilon)^2)/(2(1+2epsilon)))^n+((4+(1-2epsilon)^2)/(2(1-2epsilon)))^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,

`(df)/(d epsilon)(0)=0` and also

`(d^2f)/(d epsilon^2)(0)=5^(n-2)/2^(n-3) n (31 + 9 n)> 0` so `epsilon = 0` is a minimum and this result confirms `u_0=v_0=5/2` as the solution point, validating the proposition.