1
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1answer
35 views

Strong induction inequality proof

Use strong induction to prove that $$\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}\leq\frac{5}{8}-\frac{1}{n}$$ $$n\geq2$$ I'm not sure how to go about this. I used base cases n=2, and n=3 but ...
0
votes
1answer
23 views

How to use induction on this type of inequality?

Given $a_1,a_2,\ldots,a_n>0$ and $a_1+a_2+\ldots+a_n<\frac{1}{2}$, prove that $(1+a_1)(1+a_2)\ldots(1+a_n)<2$. Some of you may have already seen this inequality. I was the one who asked ...
0
votes
1answer
40 views

Induction proof that $4^n > 3^n+2^n$ for $n\ge2$

This is a problem with induction and proofs but I'm not sure how to start with proving this one. $$\text{Show that for any $n \geq 2$, $4^n > 3^n+2^n$}$$
5
votes
1answer
29 views

How to prove an inequality

$a$, $b$, $c$, $d$ are rational numbers and all $> 0$. $\max \left\{\dfrac{a}{b} , \dfrac{c}{d}\right\} \geq \dfrac{a+c}{b+d}\geq \min \left\{\dfrac{a}{b} , \dfrac{c}{d}\right\}$ Hope someone ...
1
vote
4answers
176 views

Induction proof of $n^{(n+1) }> n(n+1)^{(n-1)}$

The question statement from my homework booklet goes: Prove by mathematical induction that $n^{n+1} > n(n+1)^{n-1}$ is true for all integers $n \geq 2$. I've managed to come up with this ...
2
votes
1answer
125 views

Proving that $\,\sqrt [n] n < 1 + \sqrt{\frac{2}{n}}\,$ for all positive $n$

Hello I am having difficulty proving the following inequality: $$ \sqrt[n]{n} < 1 + \sqrt{\frac{2}{n}} \quad \text{for all positive integers}\,\,\, n. $$ I am trying to use mathematical induction ...
0
votes
2answers
53 views

Using induction to prove $2^{n-1}(1 + a_1a_2\ldots a_n) \geq (1+a_1)(1+a_2)\ldots(1+a_n)$ for $a_i \geq 1$

Hello I have been blasting at this inequality proof and it is just not doing what I want it to do: Prove that $2^{n-1}(a_1a_2\ldots a_n + 1) \geq (1+a_1)(1+a_2)\ldots(1+a_n)$ assuming that ...
2
votes
1answer
30 views

Inductive proof of an inequality

I am trying to prove this inequality by induction: For all $x$ in the interval $x\in [0, \pi]$, prove that: $$ |\sin (nx)| \leq n\sin(x) \textit{, n a nonnegative integer}$$ The base case is ...
3
votes
3answers
121 views

Prove that, for any positive integer n: $(a + b)^{n} \leq 2^{n-1}(a^{n}+b^{n})$

Prove that, for any positive integer n: $(a + b)^{n} \leq 2^{n-1}(a^{n}+b^{n}) $ I tried induction theorem, when $n = 1$ it is obviously right. But, say $n=k$, It does not make sense since I cannot ...
1
vote
5answers
357 views

Invalid induction proof?

Prove the following using mathematical induction. If $a_{1}, a_{1}, ... , a_{n}$ are positive real numbers such that if $$a_{1}a_{2}...a_{n} = 1 $$ then $$a_{1}+a_{2}+ ... + a_{n} \geq n$$ My proof: ...
1
vote
1answer
63 views

Stuck on inductive step: $2^x > x^n$ when $x\rightarrow \infty$

I want to show that $2^x > x^n$ when $x \rightarrow \infty$ for all $n \in \mathbb{N}$. I'm trying to do it by induction over $n$. The base case, $n = 1$, is true: $2^x > x$ when $x \rightarrow ...
0
votes
1answer
67 views

Proving an inequality for a sequence by induction

I'm having some trouble with the following problem: Let $a_n$ be a sequence defined iteratively for $n \geq 0$ as follows: $a_n = a_{m+1} + 2a_m + a_{n-m-1} + 2$ where $m$ is defined as ...
3
votes
1answer
155 views

Proof by induction - correct inductive step?

The problem: $$ x_1 \geq x_2 \geq ... \geq x_{3n} \geq x_{3n+1} \geq 0 $$ Show that: $$ x_1^2 - x_2^2 + ... - x_{3n}^2 + x_{3n+1}^2 \geq (x_1 - x_2 + ... - x_{3n} + x_{3n+1})^2 $$ I'm trying to ...
3
votes
2answers
81 views

If I'm asked to prove that $n \le m$, is it sufficient to show that $n < m$?

I have a homework question, which is to prove by induction that $\sum\limits_{r=1}^{n} \frac{1}{\sqrt{r}} \leq 2\sqrt{n}$ for every integer $n \geq 1$. I've managed to show by induction that ...
0
votes
1answer
48 views

An inequality by induction

I am reading Arthur Engel's Problem Solving Strategies. Section 8. The Induction Principle Problem 24 with its solution are attached . I do not understand the second inequality in the solution on ...
2
votes
6answers
255 views

If $a_1,a_2,\ldots,a_n>0$ and $a_1+a_2+\ldots+a_n<\frac{1}{2}$, prove that $(1+a_1)(1+a_2)\ldots(1+a_n)<2$.

If $a_1,a_2,\ldots,a_n>0$ and $a_1+a_2+\cdots+a_n<\frac{1}{2}$, prove that $$(1+a_1)(1+a_2)\cdots(1+a_n)<2$$ I've tried using Holder's inequality (the same result can easily be derived using ...
0
votes
3answers
51 views

How to show $((k+1)!)^2 2^k \leq (2(k+1))!$

How do you show that $((k+1)!)^2 2^{k+1} \leq (2(k+1))!$ This is part of an induction proof and I have not made any progress.
0
votes
1answer
42 views

Prove the following inequality using induction: $(1 + \epsilon)^n \leq 1+ (2^n - 1)\epsilon$ for every $n \geq 1$ and $0 \leq \epsilon \leq 1$

Prove the following inequality using induction: $$(1 + \epsilon)^n \leq 1+ (2^n - 1)\epsilon$$ for every $n \in \mathbb{N}: n \geq 1$ and $0 \leq \epsilon \leq 1$ I'm familiar with the concept of ...
3
votes
2answers
137 views

Induction: show that $\sum\limits_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}$ for all n $\in Z_+$

The question: show by using induction that $\sum\limits_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}$ for all n $\in Z_+$ My attempt at a solution: The base case $n = 1$ is true. First we use the ...
2
votes
3answers
68 views

Prove $2^n\cdot n! ≤ (n+1)^n$ by induction.

An induction I'm struggling with. Prove $2^n\cdot n! ≤ (n+1)^n$ by induction. An idea was to show that $2^n\cdot n! ≤ 1+n^2$ since $1+n^2 ≤ (n+1)^n$ using Bernoulli. However the inequality is ...
1
vote
1answer
38 views

Proving $\prod \limits_{k=0}^{n}(1-a_k) \geq1- \sum\limits_{k=0}^{n}a_k$

Let $(a_n)_{n \in \mathbb{N}}$ a sequence of real numbers with $0 \leq a_n \leq 1$ for all $n \in \mathbb{N}$. I want to prove the following inequality using mathematical induction: $\prod ...
1
vote
4answers
144 views

Prove using induction : $n < 3^n$

$ p(n) = n < 3^n = q(n) $ when $n=1$, $p(n)=1< 3=q(n)$ Assume the result is true for $n=m$ $p(m)=m < 3^m$ when $n = m+1$ $p(m+1) = m+1 < 3^m +1<3*3^m = 3^{m+1}=q(m+1)$ is this ...
0
votes
4answers
63 views

Proving $\displaystyle\sum_{k=1}^{m+1} \frac{1}{\sqrt{k}}\gt\sqrt{m+1}$

well the original problem was to prove the sum of k to the negative one half was more that the square root of n but it thought it would be best to use induction and get the equation displayed above. I ...
-1
votes
2answers
46 views

Mathemathical Induction Inequality

I can't do this question please help me. Prove $3^n>n^2$ for $n\ge 2$ (and also for n=0 and 1)
0
votes
0answers
38 views

Inequality sine power series (induction)

How can we show, for $k\geq 1$ and $x \geq 0$, the inequality below by induction? $\displaystyle \sin x \geq \sum_{n=1}^{2k} (-1)^{n+1} \frac 1{ (2n - 1)! }x^{2n-1} $ The base case $k = 1$ gives ...
2
votes
0answers
70 views

Inequality sine power series

How can we show, for $k\geq 1$ and $x \geq 0$, the inequality below by induction? $\displaystyle \sin x \geq \sum_{n=1}^{2k} (-1)^{n+1} \frac 1{ (2n - 1)! }x^{2n-1} $ The base case $k = 1$ gives ...
4
votes
4answers
71 views

Proving a relation with induction

I have a problem: Let $p_n$ be the $n:th$ prime number ($p_1=2, p_2=3, p_3=5$ and so on). With induction, show that $p_{n+2}>3n$ for each integer $n\geq1$. I can't figure this out because the ...
3
votes
3answers
51 views

Prove inequality $n<3^n$ using mathematical induction

Prove that $n<3^n$ where $n \in \mathbb N$, when $ n=1 $, I have proved it's true. And assumed when $n = p$ , $p<3^p$ is true. Can any body help me in showing that it is true for $n =p+1$
3
votes
2answers
81 views

Is this how you prove by induction for inequalities?

the question is here: http://cpsc.ualr.edu/srini/DM/chapters/examples/ex2.3.2.html My solution is as below:
5
votes
2answers
124 views

Math Induction Proof: $(1+\frac1n)^n < n$

So I have to prove: For each natural number greater than or equal to 3, $$(1+\frac1n)^n<n$$ My work: Basis step: $n=3$ $$\left(1+\frac13\right)^3<3$$ $$\left(\frac43\right)^3<3$$ ...
3
votes
2answers
77 views

Prove that $\sum_{k=0}^n\frac{1}{k!}\geq \left(1+\frac{1}{n}\right)^n$ [duplicate]

It basically says it all in the title. I tried solving the inequality using the bernoulli inequality somehow $$\dfrac{\displaystyle\sum_{k=0}^n\frac{1}{k!}}{(1+\frac{1}{n})^n}\geq 1,$$ but the ...
2
votes
3answers
125 views

Induction: $\sqrt{2\sqrt{3\sqrt{\cdots\sqrt n}}} < 3$ [duplicate]

I'm trying to prove that $$ \sqrt{2\sqrt{3\sqrt{4\cdots\sqrt{n}}}} < 3 $$ for any $n$ and have decided to use strong induction and instead just show that $$ \sqrt{k\sqrt{(k + 1)\cdots\sqrt{n}}} ...
3
votes
3answers
73 views

How to prove the inequality $\sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2} $ for $n\in\mathbb{Z}^+$?

I have to prove this inequality: $$ \forall n \in Z^+, \sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2} $$ So far, I have done the base cases and assumed the inequality is true for some ...
0
votes
1answer
36 views

Generalized Bernoulli's inequality

I was able to prove Bernoulli's inequality, easily by simple induction. However, I'm not sure how to prove the generalized inequality (generalized = for each sequence of numbers i=1..n): ...
2
votes
2answers
109 views

Prove by mathematical induction: $n < 2^n$

Step 1: prove for $n = 1$ 1 < 2 Step 2: $n+1 < 2 \cdot 2^n$ $n < 2 \cdot 2^n - 1$ $n < 2^n + 2^n - 1$ The function $2^n + 2^n - 1$ is surely higher than $2^n - 1$ so if $n < ...
2
votes
3answers
155 views

Proving that $n!≤((n+1)/2)^n$ by induction

I'm new to inequalities in mathematical induction and don't know how to proceed further. So far I was able to do this: $V(1): 1≤1 \text{ true}$ $V(n): n!≤((n+1)/2)^n$ $V(n+1): ...
2
votes
1answer
58 views

Why are strict inequalities stronger than non-strict inequalities?

I'm working with induction proofs involving inequalities and I am encountering example proofs that wish to show things of the sort, $n!\le\ n^n$ for every positive integer. The proof given in the ...
0
votes
1answer
59 views

Inequality with a sum and factorial

For a homework assignment we have the following question that I'm stuck on. Let $ 0 \leq y \leq 1 $ be given. $\forall m \in \mathbb{N}$, define $ \displaystyle S_m(y)=\sum_{k=0}^m \binom{m}{k}y^k$. ...
3
votes
2answers
67 views

Prove through induction that $3^n > n^3$ for $n \geq 4$

I'm new to induction and have not done induction with inequalities before, so I get stuck at proving after the 3rd step. The question is: Use induction to show that $3^n > n^3$ for $n \geq ...
4
votes
4answers
180 views

Prove $3^n \ge n^3$ by induction

Yep, prove $3^n \ge n^3$, $n \in \mathbb{N}$. I can do this myself, but can't figure out any kind of "beautiful" way to do it. The way I do it is: Assume $3^n \ge n^3$ Now, $(n+1)^3 = n^3 + ...
1
vote
2answers
82 views

Induction proof $2n+1<2^n$

I struggle to proof that: $2n+1<2^n$ By using induction. The base case is for $n\ge3$. Any help will be appreciated!
5
votes
5answers
205 views

Prove $n!>n^2$ for $n>3$

I'm aware that induction is necessary. I have been stuck on this problem for a few days now. I'm having a hard time understanding how to apply the inductive hypothesis to the inequality to arrive at ...
3
votes
7answers
686 views

Prove $n^2 > (n+1)$ for all integers $n \geq 2$

I understand that I need to use induction for this, that's not a problem. I get stuck after I try to invoke the inductive hypothesis. $P_n: n^2 > n+1$... and we want to prove $P_{n+1}: (n+1)^2 ...
0
votes
2answers
150 views

Strong Induction on Inequalities

I'm asked to indicate which natural numbers $n$ each of the below inequality is true, and then I am required to prove this via induction, but I'm wondering what that means... Strong induction? ...
3
votes
3answers
209 views

Bernoulli's Inequality

I'm asked to used induction to prove Bernoulli's Inequality: If $1+x>0$, then $(1+x)^n\geq 1+nx$ for all $n\in\mathbb{N}$. This what I have so far: Let $n=1$. Then $1+x\geq 1+x$. This is true. Now ...
5
votes
2answers
615 views

Proving the AM:GM inequality

I am doing past exam papers preparing for the finals and I came across this questions about three times: Prove that: $$\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\geq \sqrt[n]{a_{1}.a_{2}...a_{n}}$$ ...
-1
votes
3answers
97 views

Prove that $n! \geq 2^{n-1}$ for $ n\geq1$ [duplicate]

Mathematical Induction:-Prove that $n! \geq 2^{(n-1)}$ for $n\geq 1$. I tried mathematical induction but could not
14
votes
12answers
2k views

Prove that $ n < 2^{n}$ for all natural numbers $n$.

Prove that $ n < 2^{n} $ for all natural numbers $n$. I tried this with induction: Inequality clearly holds when $n=1$. Supposing that when $n=k$, $k<2^{k}$. Considering $k+1 <2^{k}+1$, ...
2
votes
4answers
436 views

For what natural numbers is $n^3 < 2^n$? Prove by induction

Problem For what natural numbers is $n^3 < 2^n$? Attempt @ Solution For $n=1$, $1 < 2$ Suppose $n^3 < 2^n$ for some $n = k \ge 1$ It looks like the inequality is true for $n = 0$, $n = 1$ ...
1
vote
2answers
60 views

Is this inequality property true?

I'm having some trouble defining weather this inequality is true or not... Basically, I wanted to know if its true that if $a \geq b$ and $c \geq d \Rightarrow a + c \geq b + d$ Well, basically ...