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) > ...
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 ...
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 + ...
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 ...
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{\...
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}, ...
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 ...
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} $...
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

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