Can anyone help me out. I reach to the conclusion that all three inequalities are wrong. But a) , b) are correct.(Answer is given in NBHM site)
46266
Let $f\in C[a,b]$. Assume that $\min_{[a,b]}f=m&gt;0$ and $M=\max_{[a,b]}f$. Which one is true?  a. $$\frac{1}{M}\int_a^bf(x)dx+m\int_a^b\frac{1}{f(x)}dx\geq 2\sqrt{\frac{m}{M}}(b-a)$$  b. $$\int_a^bf(x)dx\int\frac{1}{f(x)}dx\geq (b-a)^2$$  c. $$\int f(x)dx\int_a^b\frac{1}{f(x)}dx\leq (b-a)^2$$  For the first one  $$\frac{1}{M}\int _a^bf(x)+m\int _a^b\frac{1}{f(x)}dx\geq {\frac{m}{M}}(b-a)+{\frac{m}{M}}(b-a)$$  and the last two I am not getting. How to find these?
55741
How one can prove the following.  Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds:  $(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$  WLOG one can assume that $0\le a\le b\le c$.  It is not difficult to prove the statement when $0\le a\le b\le c\le 1$ or $1\le a\le b\le c$.
54416
Disclaimer: The statement may be false, but for now I'm operating under the assumption it's true and trying to prove it.   My workings:  I got a common denominator and expressed it as:  $$\frac{ab^2 + bc^2 + ca^2}{abc} \ge 3$$ $$ab^2 + bc^2 + ca^2 \ge 3abc$$  This is where I'm stuck. I feel as if I keep encountering this same result in various different problems and always am stumped on how to show it is true.   I tried to look at similar problems and it seems AM-GM inequality is the usual line of attack, so using that I reasoned, $$a^2 + b^2 + c^2 \ge 3abc$$, and now I need to establish $$ca^2 + ab^2 + bc^2 \ge a^2 + b^2 + c^2$$  However, this doesn't seem like it'd be true for general positive $a,b,c$.
17319
please I tried to find counterexamples to see that $l^p$ is not norm with $p&lt;1$ in the triangle inequality but I have problems with convergence when I choose some successions. Thanks.
18423
So I took a calculus course and my first homework has this one. I am stuck on this one for 2 days straight. Please help me; I will appreciate some hints.   So far we learned the axioms of a field and the axioms for inequality (trichotomy, transitivity, etc.) Do I actually need to divide it into cases where both $x,y$ are with the same sign and another case where they are with different signs?  Is there an elegant and short way for this?
83074
Question: Can anyone shed some light on this formula? I can't find any information on it.  It has three corollaries that I also need to understand:
39982
I have an inequality problem which is as follow:  If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$.  I am not so good in inequalities. So, please give me some hints so that I can proceed.  Thanks.
36735
Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed. Still, I would like to know:  if it is sound, because absolute value always creates me some problem, and if there is a shorter (neater) way to get the result (maybe without using contradiction).    Proposition: $|d(x,y)-d(y,z)| \leq d(x,z)$. Proof: We proceed by cases.  Case 1: $d(x,y)-d(y,z) \geq 0$. Assume by contradiction that $d(x,y)-d(y,z) &gt; d(x,z)$. Hence, $d(x,y)&gt; d(y,z) + d(x,z)$, contradicting the triangle inequality.  Case 2: $d(x,y)-d(y,z) &lt; 0$. Assume by contradiction that $ - d(x,y) + d(y,z) &gt; d(x,z)$. Hence, $d(x,y) &lt; d(y,z) - d(x,z)$. By applying the triangle inequality, we have that $d(y,x) \geq d(y,z) + d(x,z)$, which is by symmetry equivalent to  $d(x,y) \geq d(y,z) + d(x,z)$, obtaining the desired contradiction. $\square$     As always, any feedback is welcome. Thank you for your time.
61943
Can anyone point me to any reference on the following inequality: Let $c&gt;0, a,b \in \Bbb{R}$ $$ \|a+b\|^2 \leq (1+c)\|a\|^2+(1+{\textstyle\frac{1}{c}})\|b\|^2 $$  I'm not even sure if "Bohr's Inequality" is this inequality's name, since I can't find it anywhere in the internet.  I'm having a bad day and I got stuck in this one, apparently. Any suggestions or hints on how to prove it will be appreciated.
37460
,      &lt; 2% of people in the US have a #phd   Of those, 6.5% are Hispanic   By defending my PhD yesterday, I just became part of the 0.13% in the US that is #LatinoPhD   I find this statistic hard to believe, or if it is true maybe it is intentionally represented to look like a racial problem (although in my opinion Latino is not a race). Regarding my field, almost 40% of references are from Spanish surnames, 40% Chinese, and 20% European. Based on this analogy (maybe false) I think her statement is wrong. Is there any way or any source that can prove her claims?
21599
If I know for two numbers a and b that    $${a &lt; b }$$  Then is it correct to say that    $$ a \leq b $$  I know that the second statement is true as long as the first one is. It seems OK as it is true but from the other side it seems kinda weird (to me), to say that, if you know, that a is strictly less that b.It's like you lose some extra information.
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 $k \ge 10$?
112318
what is wrong with this??? please,can anyone help?  \begin{equation} \label{probabilityy} P^{t}({E_{0})=P_{0}+(1-P_{0})P_{1}+ (1-P_{0})(1-P_{1})P_{2}+.......+P_{n}\Pi_{k_{0}}^{n-1}(1-P_{k}) \end{equation}
81550
Prove that the following inequality holds   $$a^a b^b c^c\ge (abc)^{\frac{a+b+c}{3}}$$   if $a,b,c$ are positive.   I'm not sure how to handle these kinds of powers. Are there any "famous" but not so advanced inequalities that involve this kinds of expressions?  I was trying to do something like ordering them without lost of generality because it's symmetric, but I get nowhere and I just make the thing even more complicated.  I'm asking for solution rather than hints since this is my first time encountering these kinds of problems.
76285
I am referring to this  page 198-9:    I have troubles understanding the case with unequal variance.  In general it :   or on :    Therefore if we assume unequal variance, I would think like this:  $s_{\bar{X}_T-\bar{X}_C}=\sqrt{s^2_T/n_1+s^2_C/n_2}$  Howerver, they give this for the standard error:    I have troubles with deriving these two different formulas.
49788
If $a, b \in \mathbb{C}$, then we have the standard triangle inequality for the difference:  $$||a| - |b|| \le |a - b|.$$  I am wondering if this inequality generalizes to exponents greater that one.     My question is, for $1 &lt; p &lt; \infty$, does there exists a constant $C_p$ such that $||a|^p - |b|^p| \le C_p|a - b|^p$ for all $a, b \in \mathbb{C}$?   I am aware of the "standard" triangle inequality for $p &gt; 1$:  $$|a +b|^p \le 2^{p}(|a|^p + |b|^p),$$  and that $2^p$ may not be sharpest constant possible. If I try to use this estimate to prove, say, that $|a|^p - |b|^p \le C_p |a - b|^p$, I get stuck with an extra term that I'm not sure what to do with.  $$|a|^p - |b|^p \le 2^p|a -b|^p + (2^p -1)|b|^p.$$  A resolution on this matter is greatly appreciated.
38312
Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$.  How to prove inequality $$ ab^2+bc^2+ca^2\le 4.\tag{*} $$    In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality $$ 27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**} $$ $\color{gray}{\mbox{(Without using "universal" Lagrange multipliers method).}}$  Thanks!