I ran into this problem in my real analysis class. It sounds easy enough on the surface, and maybe it is, but I can’t seem to figure out what to do inductively as the product on the LHS is throwing me a bit off. In short, HELP, please. The question is as follows: Let n ∈ N. Show that if, a1,a2,...,an are non-negative real numbers less than 1 (0 ≤ ai &lt; 1 for i = 1,...,n), the following holds:                             1−a1)(1−a2)···(1−an)≥1−(a1 +a2 +...+an)
73031
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:  $\prod \limits_{k=0}^{n}(1-a_k) \geq  1- \sum\limits_{k=0}^{n}a_k$.  It's obviously true for $n=0$.  $n \rightarrow n+1$  Assume that it's true for n. Then one has:  $\prod \limits_{k=0}^{n+1}(1-a_k) = (1- a_{n+1}) \prod \limits_{k=0}^{n}(1-a_k) \geq (1- a_{n+1}) (1- \sum\limits_{k=0}^{n}a_k) = 1- \sum\limits_{k=0}^{n}a_k - a_{n+1} + a_{n+1} \sum\limits_{k=0}^{n}a_k = 1- \sum\limits_{k=0}^{n+1}a_k + a_{n+1} \sum\limits_{k=0}^{n}a_k=...$  How to go on?
27508
I've seen a question in my math book with this explanation above it: "If the product of n positive real numbers is 1, then the sum of these numbers must be more than n". I was wondering if this is correct, and if it is, could someone give an explanation or a proof for it?
43818
I am taking a real analysis course. I have the following statement:     Prove that if  $a$ belongs to $\mathbb R$,  such that  $0\leq a &lt;  \epsilon$, for all $\epsilon &gt; 0$, then $a = 0$   I know how to prove it (using contradiction). Now, I was wondering, how I prove almost the same thing:     Prove that if  $a$ belongs to $\mathbb R$,  such that  $0\leq a \leq \epsilon$ for all $\epsilon &gt; 0$, then $a = 0$   Can someone show me how to do it? I am stuck with the equality to show a contradiction.  Note: this is not homework, just for my own knowledge.
1126
Assume that $a_1,...a_n$ are real numbers and $-1 \leq a_i\leq 1$ for $1\leq i\leq  n$. If $$a_1^3+\ldots +a_n^3=0$$ Then show that $$a_1+a_2+\ldots+a_n\le n/3$$   I just came cross this problem the other day. I totally have no idea how to prove it. Are there any hints about which theorem, principle or method to apply?
50494
Suppose, one hundred real numbers are given and their sum is 0.Then how can I prove that at least 99 of the pairwise sums of these hundred numbers are non-negative?  I tried this: Let the real numbers be $a_i$,$1\leq i\leq100$. By the given condition, $a_1+a_2+\dots +a_{100}=0$.Suppose for the sake of contradiction, that at most 98 of the pairwise sums of these given reals are non-negative.I just tried to examine the case when exactly 98 of them are non-negative. I labelled the sums $S_1$,$S_2,\dots S_{4950}$,WLOG assume that sums $S_1,S_2,\dots S_{4852}$ are negative, the rest are non-negative. $$S_1+S_2+\dots +S_{4950}=99(\sum_{i=1}^na_i)=0$$ Case I: $S_{4853}+a_{4854}+\dots +a_{4950}=0$.That simply means that $S_1+S_2+\dots +S_{4852}=0$.But, by the given condition, $S_1+S_2+\dots +S_{4852}&lt;0$,a contradiction.After that, I am not able to make any progress.  Can anyone please suggest a more efficient method to solve this problem.I feel this approach will involve enormous case work.
99756
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that    $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$     My work: $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge \dfrac{25}{2}\implies a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}+4\ge \dfrac{25}{2}$$ Now, we have $a^2+\dfrac{1}{a^2}\ge 2$ and $b^2+\dfrac{1}{b^2}\ge 2$. Here, I am stuck, I cannot use the information provided, $a+b=1$ to any use. Please help!
31355
Suppose that $a$, $b$ and $c$ are positive real numbers and that $a \leq b + c$. By cross multiplying or otherwise, show that  $$\frac{a}{1 + a} \leq \frac{b}{1 + b} + \frac{c}{1 + c}$$  Anyone able to get this to work out? I cross multiplied them but am unable to manipulate it into an expression that verifies the statement.
19128
My question is regarding the question posed at the end of the proof.  My answer is that the result does not hold for all $m \ge 7$ because when $m=7$, the result is $343 \le 128$, which is false.  Is there another way to answer this sort of question besides directly trying different values until you get a false result?  EDIT:  Another question I have regarding this solution is:  In the proof for the inductive step, we start by assuming $k \ge 10$. But along the way, the author mentions $k \ge 1$ and $k \ge 7$ to justify the inequality.  Why do we bother to do this instead of just sticking with $k \ge 10$?
70751
Let $a$, $b$ and $c$ be positive real numbers with $abc=1$. Prove that   $$ \frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} $$   for each integer $n$.   I have used Cauchy-Schwarz inequalities and Jensen inequality. But I am stuck. I need some idea and advice on this problem. Induction would be cruel.
122936
I know I have to prove both sides here so: $\implies$: Suppose $|a| &lt; b$. Since $|a|$ is a positive number, we know $b$ is a positive number greater than $|a|$:  If $a$ is positive, $a &lt; b$ If $a$ is negative, $a &lt; b$  But for $|a| &lt; b$, it has to be that $a &gt; -b$. $a &lt; b$ and $a &gt; -b$, therefore $–b &lt; a &lt; b$ is true. $\impliedby$: Suppose $-b &lt; a &lt; b$.  If $a$ is positive, then $b$ is a greater positive number, and $-b$ would be less than $a$ If $a$ is negative, then $b$ can only be a greater positive number for $-b &lt; a$ to hold  So, $b$ is a greater positive integer than $a$. Therefore, it must be that $|a| &lt; b$. This was what I came up with but it seems convoluted and kind of forced with the cases. Any suggestions?
10408
Prove that for any natural number $n&gt;1$, we have $(2n)!&lt;(n(n+1))^n$    I tried using induction, but I failed in that approach, I rather found it was untrue, but on several case testing, I found it to be true. I am out of ideas here, please help. Thank you.
122397
Donald Knuth &quot;The Art of Computer Programming&quot; Section 1.2.1 has an exercise No.2, the solution to which I don't understand. Exercise:  There must be something wrong with this proof. What is it? Theorem. Let $a$ be any positive number. For all positive integers $n$ we have $a^{n-1}=1$. Proof. If  $n = 1, a^{n-1}=a^{1-1}=a^0=1$. And by induction, assuming that the theorem is true for $1, 2, ..., n$, we have $a^{(n+1)-1}=a^n=\frac{a^{n-1}\times a^{n-1}}{a^{(n-1)-1}}=\frac{1\times1}{1}=1$; so the theorem is true for $n+1$ as well.  As far as I understand the proof by induction, the solution should be approximately following:$a^{1-1}=1$Assume that $a^{n-1}=1$;Therefore $a^{(n+1)-1}=1\times a$;So $a^{(n+1)-1}$ does not equal $a^{n-1}$ and does not equal $1$, unless $a=1$ The solution to exercise reads:  The theorem has not been proved for $n=2$; in the second part of the proof, take $n=1$; we assume there that $a^{-1}=1$. If this condition is true (so that $a=1$), the theorem is indeed valid.  I want to compare it to another, correct, proof by induction in the example from :  Adding up Odd Numbers $1+3+5+...+(2n−1)=n^2$  Show it is true for $n=1$ $1=1^2$ is True Assume it is true for $n=k$ $1+3+5+...+(2k−1) = k^2$ is True (An assumption!) Now, prove it is true for $(k+1)$ $1 + 3 + 5 + ... + (2k−1) + (2(k+1)−1) = (k+1)^2$We know that $1 + 3 + 5 + ... + (2k−1) = k^2$ (the assumption above),so we can do a replacement for all but the last term:$k^2 + (2(k+1)−1) = (k+1)^2$Now expand all terms:$k^2 + 2k + 2 − 1 = k^2 + 2k+1$And simplify:$k^2 + 2k + 1 = k^2 + 2k + 1$They are the same! So it is true.So: $1 + 3 + 5 + ... + (2(k+1)−1) = (k+1)^2$ is True   Returning to the first example, I didn't quite understand the meaning of the phrase  The theorem has not been proved for $n=2$;  In the second (correct) proof there is no proving that it holds for $n=2$, despite covering the same topic. And Knuth himself has the proof of the theorem $1+3+5+...+(2k−1) = k^2$ similar to the one above, where he too doesn't prove it for $k=2$. So what did he mean by this phrase? Does he mean that there already is an exception in the rule stated when we take arbitrary $n$, for example $n=2$, so therefore it wouldn't hold for all examples anyway? But taking arbitrary numbers and putting it into equation instead of $n$ doesn't seems to be an effective and systematic way to debunk such proofs. How would you explain what is wrong with the exercise?
913
Suppose $a_1, a_2,\dots , a_n$ are $n$ positive real numbers with $a_1a_2 \dots a_n = 1$.   Then what is the minimum value of $(1 + a_1)(1 + a_2). . .(1 + a_n)$ ?    I think $(1 + a_1)(1 + a_2). . .(1 + a_n)$ takes its minimum value when $a_1=a_2=\dots=a_n=1$ and thus the minimum value is $2^n$.  I don't know how to prove it. Please help.
3244
Supposing that a real number $c$ is given, is the following true?  "If $n^c$ is a natural number for every natural number $n$, then $c$ is a non-negative integer."  Though this seems true, I can't prove that. Can anyone help?
28684
let $a_{i},i=1,2,\cdots,n$ be postive integer ,show that $$1^{n-1}2^{n-2}\cdots (n-2)^2(n-1)|\prod_{1\le i&lt;j\le n}(a_{i}-a_{j})$$  I know this $\prod_{1\le i&lt;j\le n}(a_{i}-a_{j})$ is  Vandermonde  determinants,and I found $$1^{n-1}2^{n-2}\cdots (n-2)^2\cdot (n-1)=1!2!3!\cdots (n-1)!=\prod_{1\le i&lt;j\le n}(j-i)$$ we only prove $$\prod_{1\le i&lt;j\le n}\dfrac{a_{j}-a_{i}}{j-i}$$ is integer  maybe consider Vandermonde determinants ? But I can't prove this
25033
This may have been asked before, however I was unable to find any duplicate. This comes from  of "Mathematical Analysis: An Introduction" by Browder. Problem 14:     If $(a_n)$ is a sequence in $\mathbb R$ and $a_n &gt; 0$ for every $n$. Then show:   $$ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$$   The middle inequality is clear. However I am having a hard time showing the ones on the left and right. (It seems like the approach should be similar for each). This is homework, so it'd be great if someone could give me a hint to get started on at least one of the inequalities.  Thanks.
114307
$a, b, c$ are positive real numbers.   Prove that $$8abc\le (a+b)(b+c)(c+a)$$  I haven't really gotten anywhere, just tried to open up $(a+b)(b+c)(c+a)$ and got null.