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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 $1\leq k\leq n$ $$2n^2\leq 2n^2+k(n-k) = (n+k)(2n-k)$$ 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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