10
votes
Accepted
Without any software and approximations prove that $\sec(52^{\circ})-\cos(52^{\circ})>1$
We have that for $\theta\in(0,\pi/2)$, $f(\theta)=\sec \theta -\cos \theta$ is an increasing function and
$$\sec \theta-\cos \theta =1 \implies \cos \theta = \frac{\sqrt 5-1}2 =\frac 1\varphi$$
that ...
- 142k
7
votes
Without any software and approximations prove that $\sec(52^{\circ})-\cos(52^{\circ})>1$
For angles less than 60 degrees, $\cos(3x)$ is a decreasing function of $\cos(x)$. So apply triple angle identity twice to $\cos(52)<(\sqrt{5}-1)/2$ to get the equivalent formulation $\cos(72) > ...
- 19k
7
votes
How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$?
What follows gives a stronger result in a natural way. At the end I give the strongest possible form of the inequality.
The inequality in question can be stated in a nicer form (two symbols less) by ...
- 9,071
6
votes
How do I make this 'intuitive' argument into a rigorous proof?
Your observation that $f(x)^3 ≤ f(x)$ when $0 ≤ f(x) ≤ 1$ is good. A similar observation extends this idea to the entire range:
For all $y ≤ 1$, we have $y^3 ≤ y^3 + \frac14(1-y)(1 + 2y)^2 = \frac14 + ...
- 7,672
6
votes
Accepted
Maximum of $\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}$ for $a, b, c \ge 0$; $a+b+c=1$
We have, for all $x\in [0, 1]$,
$$\frac{1}{18}x + \frac{1}{9} - \frac{1}{x^2 - 4x + 9}
= \frac{x(x-1)^2}{18(x^2-4x+9)} \ge 0. \tag{1}$$
Using (1), we have
$$\frac{1}{a^2 - 4a + 9}
+ \frac{1}{b^2 - 4b +...
- 25.9k
5
votes
How to proove inequality a/b + b/c + c/d + d/a>= 4?
More generally, if
$x_i > 0$ and
$\prod_{i=1}^n x_i=1$
then,
using the AM-GM inequality,
$\frac1{n}\sum_{k=1}^n x_i
\ge \left(\prod_{i=1}^n x_i\right)^{1/n}
=1
$
so
$\sum_{k=1}^n x_i
\ge n$
with ...
- 103k
5
votes
Accepted
How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$?
There are lots of ways to prove this.
One way is to square both sides and
cross-multiply.
Another way is to write
$\dfrac{2n+1}{2n+2}
=1-\dfrac{1}{2n+2}
$
and
$\dfrac{\sqrt{n+1}}{\sqrt{n+2}}
=\sqrt{\...
- 103k
5
votes
If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}<x<\frac{2014}{2015}$
We have the bound
$$\sum_{n=2}^{2015}\frac{1}{n^2}>\int_2^{2016}\frac{1}{x^2}dx=\frac{1}{2}-\frac{1}{2016}>\frac{1}{2}-\frac{1}{806}=\frac{402}{806}=\frac{201}{403}$$
by thinking about the graph ...
- 16.3k
5
votes
If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}<x<\frac{2014}{2015}$
We have
$$\frac{1}{k}-\frac{1}{k+1} < \frac{1}{k^2} < \frac{1}{k-1} - \frac{1}{k}$$
and so (telescoping sum)
$$\frac{1}{2} - \frac{1}{2016} < \sum_{k=2}^{2015} \frac{1}{k^2} < \frac{1}{1} -...
- 48.5k
4
votes
Show that $a\le|a|$
It is almost valid. In your second case, you state "Since $ a < 0$, $-a = a$" which is wrong. If $-a = a$, then $a = 0$.
You could just say that if $a < 0$, as $0 \leq |a|$, then $a &...
- 157
4
votes
How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$?
Let $f(x)= \frac{2n +x}{\sqrt{n+x}}$ and, for $x>0$, evaluate
$$f’(x) = \frac x{2(n+x)^{3/2}}\ge 0
$$
So, $f(x)$ is an non-decreasing function of $x$, which gives
$f(1)\le f(2)$ and equivalently
$$\...
- 74.9k
4
votes
How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$?
Although this is a little indirect, it keeps calculation to a
minimum.
For any positive integer $m,$ we have
$$
(m - 1)(m + 1) = m^2 - 1 < m^2,
$$
therefore
$$
\frac{m - 1}{m^2} < \frac1{m + 1},
...
- 11k
4
votes
Accepted
How to show that $\left| \frac{-x^2y-y^3-y}{x^2+y^2} \right| < \pi$?
Alright so we want to show that
$$\left\lvert \frac{-x^2y-y^3-y}{x^2+y^2} \right\rvert < \pi$$
whenever
$$1<\frac{1}{x^2+y^2}<4.$$
Now notice that then
$$\left\lvert \frac{-x^2y-y^3-y}{x^2+y^...
- 4,618
4
votes
Accepted
Prove that $\int_0^\pi xa^{\sin x}dx\cdot\int_0^\pi a^{-\sin x}dx\geq\frac{\pi^3}{2}$ for $a > 0$
We have
$$\int_0^\pi x a^{\sin x}\,\mathrm{d} x =
\int_0^{\pi/2} x a^{\sin x}\,\mathrm{d} x + \int_{\pi/2}^\pi x a^{\sin x}\,\mathrm{d} x =
\pi \int_0^{\pi/2} a^{\sin x}\,\mathrm{d} x$$
and
$$\int_0^...
- 25.9k
4
votes
$n^3 \le (\sum_{i=1}^{n} x_{i}^{2})(\sum_{i=1}^{n} \frac{1}{x_i})^2$
A second proof using AM-GM inequality two times:
$a)$ $x_1^2+x_2^2+\cdots + x_n^2 \ge n\sqrt[n]{x_1^2x_2^2\cdots x_n^2}$.
$b)$ $\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\cdots +\dfrac{1}{x_n}\right)^2 \ge \...
- 4,244
3
votes
Accepted
Solving Inequalities - a curious observation
I'm a little unclear about your question, as it mentions two critical values (I'm assuming you mean restrictions or non-permissible values) but the inequality you included only has one. However, here ...
- 2,519
3
votes
How to show that $\left| \frac{-x^2y-y^3+y}{x^2+y^2} \right| < \pi$?
$\dfrac{-x^2y-y^3+y}{x^2+y^2} = \dfrac{y-y(x^2+y^2)}{x^2+y^2}= y\left(\dfrac{1}{x^2+y^2} - 1\right) $
$0 \lt \left(\dfrac{1}{x^2+y^2} - 1\right) \lt 3 $
$1/4 \lt x^2+y^2 \lt 1 \implies 0 \leq |y| \lt ...
- 683
3
votes
How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$?
As noticed by @marty cohen in his answer and @Kurt G. in the comments, the more natural and simple way in this case is by algebraic manipulation, since all terms are positive we have
$$
\require{...
- 142k
3
votes
Maximum of $\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}$ for $a, b, c \ge 0$; $a+b+c=1$
This is something of a brute force method.
Let $f(a,b,c)=\sum_{a,b,c}1/ (a^2-4a+9)$ and $g(a,b,c)=a+b+c-1$ then calculate $\nabla f-\lambda \nabla g=0$ to find
$$
{2a-4\over (a^2-4a+9)^2}-\lambda = 0
$...
- 10.4k
2
votes
Accepted
Find two ratios given an inequality about a triangle with integer coordinates
This problem is, in fact, about inequalities and the discreteness of integers besides the basic knowledge of geometry.
$$\begin{aligned}&\quad\quad1\\
&>(|AB|+|BC|)^2-8\cdot\text{Area}(\...
- 3,328
2
votes
Without any software and approximations prove that $\sec(52^{\circ})-\cos(52^{\circ})>1$
Update:
Remarks: We can calculate $\cos (3 \cdot 52^\circ)$ in radical form. Then use $\cos 3u = 4\cos^3 u - 3\cos u$ to prove $\cos 52^\circ < \frac{\sqrt 5 - 1}{2}$. This is based on @eyeballfrog'...
- 25.9k
2
votes
Prove by induction the inequality $2^{n!}>n^n$ for $n>2$
We're trying to prove $2^{n!} > n^n$. You've shown that the base case holds - great.
In an induction proof you assume that the statement holds for an arbitrary $k$, and then you have to show it ...
- 181
2
votes
prove $\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n}-\frac{1}{n^2}$
Fact 1: Let $n \ge 2$. Let $x_1, x_2, \cdots, x_n \ge 0$ and $x_1 > 0$. Then
$$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j} \ge \frac{2n - 1}{n^2}(x_1 + x_2 + \cdots + x_n).$$
(The proof is given ...
- 25.9k
2
votes
Accepted
$\left(\int_E\sin x \,\mathrm{d}x\right)^2+\left(\int_E\cos x \,\mathrm{d}x\right)^2 \leqslant4\;?$
Define $\phi(x)=\mathbb{1}_E(x)$ for $0\leq x<2\pi$; extend $\phi$ to $\mathbb{R}$ as a $2\pi$-periodic function.
Let $z=\int_E e^{ix}\,dx$. The quantity of interest is $|z|$. Notice that $|z|=e^{...
- 27.6k
2
votes
Prove $\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1$ for $x,y,z > 0 $
Hint :
Can you show ? For $a,b,c>0$
$$\frac{1+ba+bc}{(1+a+c)^{2}}-\frac{b^{2}+ba+bc}{(b+a+c)^{2}}=\frac{\left(b-1\right)^{2}\left(a+c\right)}{\left(a+c+1\right)^{2}\left(a+b+c\right)}\geq 0$$
- 3,111
2
votes
Accepted
For which values of $s>0$ the inequality $x^{s+1}\le (y+z)$ holds true with $1\le x\le y+z$?
For $x=1$ the inequality always holds. For $x>1$ we have
\begin{align}
x^{s+1} \le w & \iff (s+1)\log x\le \log w\\
&\iff s+1 \le \frac{\log w}{\log x}\\
&\iff s \le \frac{\log w}{\log ...
- 13.5k
2
votes
Accepted
Let $x \in R$. Then, prove that $x^2+|x-6|>5$
For your case $(3)$, notice that if $x<6$ then $|x-6|=6-x$. Then, your inequality becomes
$$x^2+6-x>5$$
$$x^2-x+1>0$$
$$(x-\frac{1}{2})^2+\frac{3}{4}>0$$
which is true.
- 1,103
2
votes
Prove that $(a-d)^2+(b-c)^2\geq 1.6$ if $a^2+4b^2=4$ and $cd=4$
Using $a^2 + 4b^2 = 4$ and $cd = 4$, we have
\begin{align*}
(a-d)^2 + (b-c)^2 &= (a-d)^2 + (b - c)^2 + \frac12 (a^2 + 4b^2 - 4)\\[5pt]
&= \frac16(3a - 2d)^2 + \frac13(3b - c)^2 + \frac23c^2 +...
- 25.9k
2
votes
Accepted
prove that there exist $x_1\neq x_2, x_1,x_2\in [0,1]$ so that $\int_{x_1}^{x_2} f(x)dx = (x_1-x_2)^{2002}$
You're right that the IVT is useful here: Wlog we can assume $a < b$ and $c < d$, now define
$$
g: [0, 1] \to \mathbb{R}\\
t \mapsto |F(ta+(1-t)c) - F(tb+(1-t)d)| - |t(a-b) + (1-t)(c-d)|^{2002}
$...
- 351
2
votes
How to show that $\left| \frac{-x^2y-y^3-y}{x^2+y^2} \right| < \pi$?
The problem with your solution is that by bounding separately the two expression and then putting them togheter we are not able to obtain a "good" estimation for the upper bound.
For example ...
- 142k
Only top scored, non community-wiki answers of a minimum length are eligible