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

Question

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

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

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Answer 1 · 2017-03-12T22:58:23

See below.

Explanation:

This problem can be stated as a minimization problem.

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

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 · 2017-03-14T00:14:12

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.

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