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)#

2 Answers
Mar 12, 2017

Answer:

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#

Mar 14, 2017

Answer:

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.