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I have to show that a_n >= b_n for N{0}, Knowing that a > 0 and b > 0. See the picture

enter image description here

How can I show the inequality with mathematica? Here is what I tried, but it didn't work (I began Mathematica one day ago)

enter image description here

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Prove the induction step. This will allow us to know that no matter the n, so long as a[n]>0 and b[n]>0, then a[n+1] >= b[n+1] by substituting in their definitions.

FullSimplify[(a[n] + b[n])/2 >= 2/(1/a[n] + 1/b[n]), {a[n] > 0, b[n] > 0}]

True

We should also prove that the precondition of a[n]>0 and b[n]>0 implies that a[n+1]>0 and b[n+1]>0, just so that we don't have any surprises during induction.

FullSimplify[(a[n] + b[n])/2 > 0, {a[n] > 0, b[n] > 0}]
FullSimplify[2/(1/a[n] + 1/b[n]) > 0, {a[n] > 0, b[n] > 0}]

True

True

Then by considering the condition that a[0]>0 and b[0]>0, the application of induction shows that for all $n\in\mathbb{N}$ that $a_{n+1} \ge b_{n+1}$. Thus, so long as $a_0\ge b_0$ or $n\ge 1$, $a_n\ge b_n$.

Note that $a_0$ may be less than $b_0$, however.

Showing this directly in Mathematica is a bit trickier, however.

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  • $\begingroup$ This relation is true in both situations: a>b , b>a, a=b. Maybe I can do your first step, and then proove that a[n] is > 0 for every n, and same for b[n] ?? $\endgroup$ – nolw38 Feb 20 '18 at 7:55
  • $\begingroup$ @nolw38: So long as $n \ge 1$, $a_n \ge b_n$, regardless. But $a_0$ and $b_0$ are independent, that's all I was trying to point out. $\endgroup$ – eyorble Feb 20 '18 at 7:56
  • $\begingroup$ Ok Thanks and sorry! As I am new in Mathematica, can you explain to me the utility of FullSimplify? $\endgroup$ – nolw38 Feb 20 '18 at 7:58
  • $\begingroup$ @nolw38 In this case, I am using FullSimplify to reduce mathematical expressions to the simplest forms Mathematica can find. The first argument is the expression to be simplified, while the second argument is the assumptions that can be made while simplifying. In this case, all 3 of these statements simplify to True, meaning that so long as the assumptions are satisfied, they are always true. $\endgroup$ – eyorble Feb 20 '18 at 7:59
  • $\begingroup$ Ok I understand! So, to prove than a[n] and b[n] are > 0, for n > 0, is there a simple way to show it? BTW, why didn't it work with RSolve? $\endgroup$ – nolw38 Feb 20 '18 at 8:11

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