7
votes
Let $x,y,z\in[0,1]$. Find the maximum value of $\sqrt{|x-y|}+\sqrt{|y-z|}+\sqrt{|z-x|}$.
WLOG fix $x\geq y\geq z$. Then our expression $E$ is
$$E=\sqrt{x-y}+\sqrt{y-z}+\sqrt{x-z}$$
Now, note that as $x$ increases, so does $E$ and so for the maximum value of $E$ we should have $x=1$. ...
- 3,221
6
votes
An interesting recurrent equality, possibly easier to solve in its differential form?
Your question comes in two variants. One is discrete.
The other continuous.
The discrete variant asks about the equation
$$ f(n+1) - f(n) = c f(n)\sum_{m=0}^n f(m). \tag{1} $$
A nicer version of this ...
- 29.3k
6
votes
Accepted
Elementary proof that $\text{Re}\big(\frac{z^{n+1} - n z - z + n}{(z-1)^2}\big) \ge \frac{n}2$?
This works by direct computation:
$h(z)=f(z)-\frac{n}{2}=\frac{z^{n+1} - n z - z + n}{(z-1)^2}-\frac{n}{2}=\frac{n-z-z^2-..-z^n}{1-z}-\frac{n}{2}$ so the sign of $\Re h(z)$ is the sign of $$\Re [2(1-\...
- 20.7k
6
votes
Accepted
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$
We first prove the following lemma.
Lemma. Let $\lambda>0$. Then
$$\sup_{x>0}\left(\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+\frac{\lambda}x}}\right)=\begin{cases}\frac2{\sqrt{1+\sqrt\lambda}}&\text{...
- 26.2k
5
votes
Accepted
Is there a shorter or more trivial way to prove that $ x > \cos (x)-\cos (2 x) $ holds for all $x>0$?
This can be seen geometrically.
Let $P = (\cos x, \sin x)$ and $Q = (\cos 2x, \sin 2x)$; then arc $PQ$ on the unit circle has length $x$. Its projection onto the horizontal axis is an interval ...
- 113k
5
votes
Is there a shorter or more trivial way to prove that $ x > \cos (x)-\cos (2 x) $ holds for all $x>0$?
If $x>2$, the statement is trivial. For $0<x\leq 2,$ one has $$x>2\sin\left(\frac x2\right)\geq 2\sin\left(\frac{3x}2\right)\sin\left(\frac x2\right)=\cos(x)-\cos(2x).$$
- 6,189
4
votes
Find number of continuous functions satisfying the equation $4\int_{0}^{\frac{3}{2}}f(x)dx+125\int_{0}^{\frac{3}{2}}\frac{dx}{\sqrt{f(x)+x^2}}=108$
Let $g(x)=f(x)+x^2$ first of all. Also note that $g(x)\geq 0$. Now,
$$4\int_0^{3/2}g(x)\textrm dx+125\int_0^{3/2}\dfrac{1}{2\sqrt{g(x)}}\textrm dx+125\int_0^{3/2}\dfrac{1}{2\sqrt{g(x)}}\textrm dx$$
$$\...
- 3,221
4
votes
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$
Since the inequality is both homogeneous and cyclic, assume that $a = \max(a, b, c, d) = 1$.
It suffices to prove that, for all $b, c, d \in (0, 1]$,
$$\frac{1}{\sqrt{1+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\...
- 22.6k
3
votes
Accepted
Inequality involving integral of 1/log(t)
By substituting $u = \frac{\log t}{\log x}$, we find that
$$ \int_{x}^{x^2} \frac{1}{\log t} \, \mathrm{d}t =\int_{1}^{2} \frac{x^u}{u} \, \mathrm{d}u. $$
Now use the fact that $x \leq x^u \leq x^2$ ...
- 140k
3
votes
Accepted
Find the value $\hat{\beta}$ which minimizes $\sum_{i=1}^{4}|i||i- \beta|$ for $\beta \in \mathbb{R}$
For this specific case: the function is linear on each of the intervals $[-\infty, 1], [1, 2], [2, 3], [3, 4], [4, \infty]$. Thus, if it achieves a minimum it will do so at one of the integers $1, 2, ...
- 24.9k
3
votes
Is there a shorter or more trivial way to prove that $ x > \cos (x)-\cos (2 x) $ holds for all $x>0$?
Almost solution.
Mean value theorem. Note $\cos' = -\sin$. For $x>0$ we have $ 0 < x < 2x$ and
$$
\frac{\cos(2x) - \cos(x)}{2x-x} = -\sin(\xi)
$$
for some $\xi$ between $x$ and $2x$. Thus
$...
- 96.5k
3
votes
Is there a shorter or more trivial way to prove that $ x > \cos (x)-\cos (2 x) $ holds for all $x>0$?
For $x > 0$ is
$$
\cos(x)-\cos(2x) = \int_x^{2x} \sin(t) \, dt < \int_x^{2x} 1 \, dt = x \, .
$$
Strict inequality between the integrals holds because $\sin(t) < 1$ for “almost all” $t$.
- 86.9k
3
votes
Accepted
An unusual inequality problem involving real exponential power
The form $4\omega^\alpha+\omega^{n-4\alpha}$ strongly suggests it's AGM time:
$$
\frac{4\omega^\alpha+\omega^{n-4\alpha}}{5}\ge\sqrt[5]{\omega^{4\alpha}\omega^{n-4\alpha}} = \omega^{n/5}.
$$
So we ...
- 17k
2
votes
Prove $\sqrt{x-1} + \sqrt{y-1} \le xy, x \ge 1, y \ge 1$
Using Cauchy Schwarz and AM-GM
$$\sqrt{x-1}+\sqrt{y-1} \le \sqrt{2(x+y-2)} \le \frac{2(x+y-2)+1}{2} \le x+y-1 $$
Now it suffices to prove that
$$x+y-1 \le xy$$
Wich is equivalent to $$(x-1)(y-1) \ge 0$...
- 86
2
votes
Solve: $\sec(2x) \ge\sec(x) , x\in [0,\pi]$\ {$\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$}
Recall that $\sec(x) = \frac{1}{\cos x}$ and $\cos(2x) = 2 \cos^2(x) - 1$. So, in terms of $\cos(x)$, you have:
$$\frac{1}{2 \cos^2(x) - 1} \ge \frac{1}{\cos(x)}$$
For convenience, let $c = \cos(x)$.
...
- 7,679
2
votes
Accepted
How to solve (not numerically) the inequality $x^a-x^b -c>0$?
Assuming $b >a>0$, let $x^a=y$, $k=\frac b a>1$ and the problem becomes
$$f(y)=y-y^k-c >0$$ The first derivative cancels at
$$y_*=k^{-\frac{1}{k-1}} $$ At this point $$ f(y_*)=-c+(k-1)\, ...
- 214k
2
votes
Find the value $\hat{\beta}$ which minimizes $\sum_{i=1}^{4}|i||i- \beta|$ for $\beta \in \mathbb{R}$
The right part, after division by $10$, can be viewed as expectation of absolute difference of random variable $i$ and constant $\beta$, but with not uniform distribution on $i$ (we have $P(i = k) = \...
- 7,921
2
votes
Accepted
Clarkson inequality for complex numbers
Let's assume for simplicity that $a\neq 0$ and $b\neq 0.$ Then $$\displaylines{|a+b|^2=|a|^2+|b^2|+2\Re (a\bar b)=
|a|^2+|b^2|+2|a|\,|b|\cos t\\ |a-b|^2=|a|^2+|b^2|-2\Re (a\bar b)=
|a|^2+|b^2|-2|a|\,|...
- 4,937
2
votes
Accepted
De Bruijn's inequality for series
I do not have a reference, but here is a possible proof:
For positive integers $N, n$ and $j \in \{ 1, \ldots, k \}$ holds the following implication:
$$
f(na_j) > f(N) \implies na_j < N \...
- 86.9k
2
votes
How do I prove that $\forall a \forall b(\neg a<b\iff b\leq a)$
Assuming we defined the natural numbers with PA
We proceed via induction on a:
Let $a = 0$:
Case 1: $b=0$:
Prove that the $0<0$ is false (fill in the details) then prove that $0 \leq 0$ is true
...
- 174
2
votes
Accepted
Is that quantity bounded from below?
Assuming $N$ to be independent of $t$, for $p=p'=2$ and $F(t)=t^2/2+t^4e^{t^2}$ the quantity in question becomes
$$e^{t^2}t^2/(s+1)-2N(t^2/2+t^4e^{t^2})$$
which is unbounded from below.
- 16.8k
1
vote
Accepted
Simple word problem about inequalities
If A can do the work in $x$ seconds, then the fraction of the total work A finishes within one second is $\frac{1}{x}$. Similarly, we obtain for B $\frac{1}{x+2}$.
Since you're interested in B's ...
- 698
1
vote
Accepted
An inequality: from the complex to the real case.
$|\frac{z+w}{2}|^q+|\frac{z-w}{2}|^q =|z|^{q} \frac {(1+t)^{q}+(1-t)^{q}} {2^{q}}$ where $t=|\frac w z|$. [Note that $t=\frac w z$ or $t=-\frac w z$]. Hence, $|\frac{z+w}{2}|^q+|\frac{z-w}{2}|^q \...
- 301k
1
vote
Prove $\sqrt{x-1} + \sqrt{y-1} \le xy, x \ge 1, y \ge 1$
you can use schwartz inequality directly
we know ;$a_1 b_1+a_2b_2 \leq \sqrt{a_{1}^2+a_{2}^2} \sqrt{b_{1}^2+b_{2}^2}$
just put $a_1 =\sqrt{x-1}$
$a_2 =1$
$b_1 =1$
and
$b_2 =\sqrt{y-1}$
so you will get ...
- 40
1
vote
If $−1<x<4$ then determine $a$ and $b$ in $a<2x+3<b$.
Yes, you did do something wrong: abuse notation without understanding what is based on, and misinterpreting the result that should be obtained.
$a$ is in the interval $(-\infty,1]$ but that doesn't ...
- 2,806
1
vote
An interesting recurrent equality, possibly easier to solve in its differential form?
Being a combination of product and sum, it looks difficult
that there might be a closed form.
The best I can suggest is to make the substitution
$$
f(n) = 2^{g(n)}
$$
I am using $2$ as a base because ...
- 33.3k
1
vote
Clarkson inequality for complex numbers
Since the inequality in question is used primarily to derive bounds in $L_p$, namely Clarkson's inequality for $1<p<2$:
$$\Big\|\frac{f+g}{2}\Big\|^{p'}_p+\Big\|\frac{f-g}{2}\Big\|^{p'}_p\leq \...
- 26.4k
1
vote
Accepted
Let $X$ be an exponential random variable and $\Bbb{P}(X \in A) < \Bbb{P}(X \in B)$. Is $\Bbb{P}(aX \in A) < \Bbb{P}(aX \in B)$ for $a > 0$?
Not necessarily.
Take for instance $A=(x,+\infty)$ and $B=(0,y)$ for any $x,y>0$ such that $1-\exp(-x)>\exp(-y)$, so that $\mathbb P(X\in A)<\mathbb P(Y\in B)$. For instance, $x=\ln(3)$ and $...
- 3,256
1
vote
Bounding spectral radius of special matrix (extension of the extension)
This is not an answer, but some thoughts.
Consider the case when $D_{2}$ has only one positive diagonal entry, and all other diagonal entries equal to $0$ -- similar to what we did here.
First, as in ...
- 326
1
vote
Solution verification: is that quantity bounded from below?
The claim only holds for $N \geq 0$ and $p \geq 2$.
To see why you need $N \geq 0$, note that if $N < 0$ then
$$\frac{|t|^p}{s+1}e^{|t|^{p'}} - 2NF(t) = \frac{|t|^p}{s+1}e^{|t|^{p'}} + CF(t)$$
for $...
- 618
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