# inducation

• February 15th 2010, 03:31 AM
inducation
need help with two questions:

prove with inducation that for every natural n exists:
$\frac{1}{(1+a)^n}+\frac{1}{(1-a)^n}\geq2$
(-1<a<1)

the second:
$\frac{2}{(a+b)^n}<\frac{1}{a^n}+\frac{1}{b^n}$
(a>0, b>0)

many thanks
• February 15th 2010, 04:56 AM
anyone?
• February 15th 2010, 05:19 AM
Quote:

need help with two questions:

prove with inducation that for every natural n exists:
$\frac{1}{(1+a)^n}+\frac{1}{(1-a)^n}\geq2$
(-1<a<1)

the second:
$\frac{2}{(a+b)^n}<\frac{1}{a^n}+\frac{1}{b^n}$
(a>0, b>0)

many thanks

For the 1st one....

If the inequality is valid, then we try to use it to prove the following

$\frac{1}{(1+a)^{n+1}}+\frac{1}{(1-a)^{n+1}}\ \ge\ 2$

Proof

$\frac{1}{(1+a)^n}+\frac{1}{(1-a)^n}=\frac{1+a}{(1+a)^{n+1}}+\frac{1-a}{(1-a)^{n+1}}$

$=\frac{1}{(1+a)^{n+1}}+\frac{a}{(1+a)^{n+1}}+\frac {1}{(1-a)^{n+1}}-\frac{a}{(1-a)^{n+1}}$

$=\frac{1}{(1+a)^{n+1}}+\frac{1}{(1-a)^{n+1}}+\frac{a}{(1+a)^{n+1}}-\frac{a}{(1-a)^{n+1}}$

As this is $\ge\ 2$ then we only need ask if

$\frac{a}{(1+a)^{n+1}}-\frac{a}{(1-a)^{n+1}}\ \le\ 0$

Now, since |a|<1

$\frac{a}{(1+a)}^{n+1}<1$

$\frac{a}{(1-a)^{n+1}}>1$

Therefore, the extra terms sum to a negative value
and the overall sum is at least 2.

Therefore

$\frac{1}{(1+a)^{n+1}}+\frac{1}{(1-a)^{n+1}}\ \ge\ 2$

if $\frac{1}{(1+a)^n}+\frac{1}{(1-a)^n}$ is

Now prove for the first term
• February 15th 2010, 05:39 AM
wow that was beautiful, thank you
• February 15th 2010, 05:43 AM
For the 2nd one..

$\frac{2}{(a+b)^n} < \frac{1}{a^n}+\frac{1}{b^n}$ ?

If this is true, then the following must be true

$\frac{2}{(a+b)^{n+1}} < \frac{1}{a^{n+1}}+\frac{1}{b^{n+1}}$ .... (1)

We try to use the first statement to prove the second one..

$\frac{2}{(a+b)^n}\ \frac{1}{(a+b)} < \frac{1}{a^n}\ \frac{1}{(a+b)}+\frac{1}{b^n}\ \frac{1}{(a+b)}$

$\frac{2}{(a+b)^{n+1}} < \frac{1}{a^{n+1}+a^nb}+\frac{1}{b^{n+1}+b^na}$

Since these denominators are larger than those in (1),
then these fractions are even smaller than those in (1)

Hence, the inductive step is proven,
meaning that if the inequality holds for any n=k, it holds for the next n=k+1 and the next n=k+2 and the next n=k+3 etc etc,
therefore it only needs to be validated for an initial term.
• February 15th 2010, 05:52 AM