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## Hot answers tagged inequality

12

One series for $\log(2)$ is $$\log(2)=1-\frac12+\frac13-\frac14+\dots$$ For an alternating series, the terms of which are decreasing in absolute value, the sum is between any two consecutive partial sums, such as $$1-\frac12+\frac13=\frac56$$ and $$1-\frac12+\frac13-\frac14=\frac7{12}$$

8

Hint: Note that $$\left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right)=ab+\frac{1}{ab}+2.$$

8

Your first statement $x + \frac 1x > 2$ is not fully correct, you have $$x + \frac 1x \ge 2$$ for $x > 0$, with equality exactly for $x = 1$. And as the example $a=b=1$ shows, you can only conclude that $$a+\frac{1}{b} \ge 2 \quad \text{ or } \quad b+\frac{1}{a} \ge 2 \, .$$ which is your claim with $\ge$ instead of $>$. This follows from ...

7

Hint:$$\big(\sqrt{a+b}\big)^2 - \big(\sqrt{a}+\sqrt{b}\big)^2 = \big(a+b\big)-\big(a+b+2\sqrt{ab}\big)=-2\sqrt{ab}$$

6

You don't need induction, since it is enough to show $$n(n+2)<(n+1)^2$$

6

For natural numbers. It is clear that $3\leq p$, similarly, $4\leq n$, and $5\leq m$. Thus, $\frac 13\geq \frac 1p$, similarly, $\frac 14\geq \frac1n$, and $\frac 15\geq \frac 1m$. Adding these gives you $$\frac13+\frac14+\frac15\geq\frac 1p+\frac 1m+\frac1n$$ which gives us the final result $$\frac 1p+\frac 1m+\frac1n\leq \frac{47}{60}$$ Note that this is ...

6

Yep - triangle inequality gives that $\vert(x+y)+(-y)\vert\le\vert x+y\vert+\vert(-y)\vert$ $$\implies\vert x \vert \le \vert x+y \vert+\vert y \vert$$ $$\implies \vert x+y \vert \ge \vert x \vert - \vert y \vert$$ Perhaps worth noting that the restriction $\vert x \vert \ge \vert y \vert$ isn't needed.

6

Here's my attempt to expand and collect terms: \begin{align} \frac 13(x + y + z)^2 & \stackrel{?}{\ge} xy + xz + yz\\ x^2 + y^2 + z^2 + 2xy + 2xz + 2yz & \stackrel{?}{\ge} 3xy + 3xz + 3yz \\ x^2 + y^2 + z^2 - xy - xz - yz & \stackrel{?}{\ge} 0 \\ \frac 12(x - y)^2 + \frac 12(x - z)^2 + \frac 12(y - z)^2 & \stackrel{?}{\ge} 0. \end{align}

5

Neither of them holds, in general, for stochastic integrals. The trouble already starts if you consider measures which need not to be non-negative, i.e. signed measures. For a signed measure $\mu: (\Omega,\mathcal{A}) \to \mathbb{R}$ we cannot expect that the triangle inequality $$\left| \int f(x) \, \mu(dx) \right| \leq \int |f(x)| \, \mu(dx) \tag{1}$$ ...

5

There are positive integrals that relate $\log(2)$ to its first four convergents: $0,1,\frac{2}{3},\frac{7}{10}$. \begin{align} \int_0^1\frac{2x}{1+x^2}dx &= \log\left(2\right) \\ \int_0^1\frac{(1-x)^2}{1+x^2}dx &= 1-\log\left(2\right) \\ \int_0^1\frac{x^2(1-x)^2}{1+x^2}dx &= \log\left(2\right)-\frac{2}{3} \\ \int_0^1\frac{x^4(1-x)^2}{1+x^2}dx ... 5 Notice that |a_i|^2=a_i^2. Using the Generalized Mean Inequality, we see\frac{|a_1|+\cdots+|a_n|}{n}\leq \sqrt{\frac{|a_1|^2+\cdots+|a_n|^2}{n}}$$which we can rewrite to$$|a_1|+\cdots+|a_n|\leq \sqrt{n}\sqrt{a_1^2+\cdots+a_n^2}$$and since \sqrt{n}\leq n we get$$|a_1|+\cdots+|a_n|\leq n\sqrt{a_1^2+\cdots+a_n^2}$$Alternatively, we can use ... 5 I assume 3 \beta^2 is meant rather than 2\beta^2. This is exactly the Cauchy-Schwarz inequality for the inner product$$\langle (\alpha,\beta),(x,y) \rangle = 5\alpha x + \alpha y + \beta x + 3\beta y.$$5 Well,$$f(x,y,z)=(1-x)(1-y)(1-z)+xyz$$so since both terms are positive (x,y,z\in(0,1) so all factors of both terms are positive),$$f(x,y,z)>0$$5 Yes.$$ \left|\sqrt {a^2+x^2} - \sqrt {a^2+y^2}\right| =\frac{\lvert x^2-y^2\rvert}{\left|\sqrt {a^2+x^2} + \sqrt {a^2+y^2}\right|} = |\sqrt {x^2} - \sqrt {y^2}|\cdot \frac{|\sqrt {x^2} + \sqrt {y^2}|}{\left|\sqrt {a^2+x^2} + \sqrt {a^2+y^2}\right|} \leq |\sqrt {x^2} - \sqrt {y^2}| $$4 Your sum can be expressed as a Riemann Sum. Specifically$$\frac 1{n+1}+\dots+\frac 1{2n}<\int_n^{2n}\frac {1}{x}dx=ln(2n)-ln(n)=ln(2)\sim.693$$4 Using 2 a * b \le a^2 + b^2  to prove$$\frac{2n + 1}{2} = \frac{ n + (n+1)}{2} \ge \sqrt{n} \sqrt{n + 1} $$Divide both sides by n+1$$\frac{2n+1}{2n+2} \ge \frac{\sqrt{n}}{\sqrt{n+1}}$$4 Hint: if a=\cos\alpha, \sqrt{1-a^2}=\sin\alpha, b=\cos\beta and \sqrt{1-b^2}=\sin\beta, then you need to show that$$ \lvert\cos\alpha\sin \beta+\cos \beta\sin\alpha-\sqrt3 \sin\alpha\sin\beta+\sqrt3 \cos\alpha\cos\beta| =\lvert\sin(\alpha+\beta)-\sqrt3\cos(\alpha+\beta)|\le2, $$that is,$$ ...

4

I thought it might be instructive to present a way forward that does not rely on calculus, but rather elementary analysis only. In THIS ANSWER and THIS ONE, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality $$e^x\ge 1+x \tag 1$$ Setting $x=-z/(z+1)$ into ...

3

If you multiply both sides by $6$ and then expand the square, you get $$2x^2+2y^2+2z^2+4xy+4yz+4xz\geq 6xy+6yz+6xz,$$ which you can rewrite as $$(x^2+y^2)+(y^2+z^2)+(x^2+z^2)\geq 2xy+2yz+2xz.$$ Then rewrite this as $$(x-y)^2+(y-z)^2+(x-z)^2\geq 0.$$

3

J.G.'s elegant answer deserves to be restated in the following simple (and perhaps more familiar) form: we can approximate the integral $$\ln 2 = \int_2^4 \frac1x\, dx$$ by a Riemann sum splitting the interval [2,4] into two parts of width $1$. Since $1/x$ is strictly decreasing on this interval, the left-Riemann sum is a (strict) overestimate, and the ...

3

To the extent that arithmetic is part of real analysis, it suffices to note that $$3^7=2187\lt4096=2^{12}$$ and $$2^6\cdot4^5=2^{16}=65536\lt100000=10^5=2.5^5\cdot4^5$$ The desired inequalities now follow from the fact that $2.5\lt e\lt3$.

3

The reverse inequality is true for $n≥1$. If $1≤k≤n$ then $\frac{1}{n+k}≥\frac{1}{n+n}$. Therfore, you get :

3

Expanding on the idea of the other answer: $$\frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}=\frac1n\left(\frac1{1+\frac1n}+\frac1{1+\frac2n}+\ldots+\frac1{1+\frac nn}\right)\rightarrow$$ $$\rightarrow\int_0^1\frac{dx}{1+x}=\log2<\frac34$$

3

You don't need calculus here (although it would make things easier). Let $$S=\frac{1}{n+1}+\frac {1}{2n}+\frac{1}{n+3}+...+\frac{1}{2n}$$ and write it both ways, à la Gauss: \begin{align} 2S &= \frac{1}{n+1}+\frac {1}{n+2}\,\,\,+\,\,\frac{1}{n+3}\,+...+\,\,\,\frac{1}{2n} \\ &\,\,\,\,+\frac{1}{2n}\,\,\;+\frac ... 3 A variation on the inequality that @TonyK already mentioned. For 1\leq k\leq n2n^2\leq 2n^2+k(n-k) = (n+k)(2n-k)$$and so, dividing both sides by 2n^2(n+k),$$\frac1{n+k}\leq\frac1n-\frac{k}{2n^2}.$$Then$$\sum_{k=1}^n\frac1{n+k}\leq 1 - \sum_{k=1}^n\frac{k}{2n^2}=1-\frac{n(n+1)}{4n^2}=\frac34-\frac1{4n}.3 I can show the stronger statement {20n \choose 10n}\ge {2n \choose n}^{10}. Suppose I have 20n books divided into 10 collections. The first one is the number of ways to choose 10n books out of my 20n books. The second one is the number of ways to choose n books out of my 2n books for every collection. The first one is larger, since there if ... 3 The sum can be written as \begin{align} \frac{1}{n+1} + \frac{1}{n+3} + \ldots + \frac{1}{3n - 1} & = \sum_{i=1}^n \frac{1}{n + 2i - 1}. \end{align} Now recall the AM-HM inequality: \frac 1n\sum_{i=1}^n(n + 2i - 1) > \frac{n}{\sum_{i=1}^n \frac{1}{n + 2i - 1}}. $$(The requirement that n > 1 guarantees that the inequality is strict.) ... 2 f(x) = 1/x is strictly convex, therefore$$ \frac{1}{2n} < \frac 12 \left( \frac{1}{n+k} + \frac{1}{3n-k} \right) $$for k = 1, ..., n-1, or$$ \frac{1}{n+k} + \frac{1}{3n-k} > \frac {1}{2n} + \frac {1}{2n} $$Combining terms pairwise from both ends of the sum shows that$$ \frac{1}{n+1} + \frac{1}{n+3}+\dots+\frac{1}{3n-3} + \frac{1}{3n-1} > ...

2

Suppose you have proved the inequality for $n=2$. Suppose it holds for $n\ge2$; then $$\def\vA{\vphantom{A}} \sqrt{x_1+\dots+x_n+x_{n+1}\vA} \le \sqrt{x_1+\dots+x_n\vA}+\sqrt{x_{n+1}\vA} \le \sqrt{x_1\vA}+\dots+\sqrt{x_n\vA}+\sqrt{x_{n+1}\vA}$$ The first $\le$ is from the case $n=2$, the second one from the induction hypothesis. For the $n=2$ case, just ...

2

Any statement that needs to be proved for all $n\in\mathbb{N}$, will need to make use of induction at some point. We have \begin{align} S_n & = \sum_{k=1}^n \dfrac1{n+2k-1} = \dfrac12 \left(\sum_{k=1}^n \dfrac1{n+2k-1} + \underbrace{\sum_{k=1}^n \dfrac1{3n-2k+1}}_{\text{Reverse the sum}}\right)\\ & = \dfrac12 \sum_{k=1}^n ...

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