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14

The expression is only defined for $x\geq 5$, however for all $x\geq 5$ the left hand side is positive and the right hand side is non-positive, hence it holds for all $x\geq 5$.


10

Expanding on comments above to give an intuitive idea of this result: write $f(x) = \frac{1}{1+x}$ and set $n = 4$. A rough diagram: Then, for example, the area of the first (left most) rectangle is $$f(1+1/4) \cdot \Delta x = \frac{1}{1+1/4} \cdot \frac{1}{4}$$ The area of all four rectangles is $$\sum_{k=1}^4 f(1 + k/4) \Delta x = \frac{1}{n} ...


9

The function $f: x \mapsto (x+\frac2x)^2$ is convex on $(0,\infty)$, so by Jensen's inequality we have $$ \frac{\left(a+\frac2a\right)^2+\left(b+\frac2b\right)^2}{2} = \frac{f(a)+f(b)}{2} \geq f\left( \frac{a+b}{2} \right) = f\left(\frac12\right) = \left( \frac92 \right)^2 = \frac{81}{4}. $$


7

Since $f$ has mean zero, $$ \int_{0}^{1} x\,f(x)\,dx = \int_{0}^{1}\left(x-\frac{1}{2}\right)\,f(x)\,dx,\tag{1}$$ and by the Cauchy-Schwarz/Buniakowski inequality we have: $$ \left(\int_{0}^{1}\left(x-\frac{1}{2}\right)\,f(x)\,dx\right)^2 \leq \int_{0}^{1}f(x)^2\,dx \int_{0}^{1}\left(x-\frac{1}{2}\right)^2\,dx \tag{2}$$ so we just have to check that: $$ ...


7

By$\let\leq\leqslant$ Cauchy-Schwarz, $$\begin{align*}(x_1y_1+\cdots+x_ny_n)^2&\leq(x_1^2y_1^2+\cdots+x_n^2y_n^2)\cdot(1^2+\cdots+1^2)\\&=n(x_1^2y_1^2+\cdots+x_n^2y_n^2)\end{align*}\tag A$$ and $$(x_1y_1+\cdots+x_ny_n)^2\leq(x_1^2+\cdots+x_n^2)(y_1^2+\cdots+y_n^2).\tag B$$ $\frac1n\cdot({\rm A})+(1-\frac1n)\cdot({\rm B})$ gives your inequality: ...


7

Note that for $x \neq 0$ and $y \neq 0$ we have $$ \left | \frac{x^2y}{x^2 + y^3} \right| = \left| \frac{y}{1 + \frac{y^3}{x^2}} \right| = \left| \frac{1}{\frac{1}{y} + \frac{y^2}{x^2}} \right| \leq \left| \frac{1}{\frac{1}{y}} \right| = \left| y \right|. $$ If $x = 0$ or $y = 0$ (but not $x = y = 0$), then we also have $$ \left | \frac{x^2y}{x^2 + y^3} ...


5

You could express this as $\log(1/2) > -\dfrac{253}{365}$. The series $$\log(1/2) = \log(1-1/2) = -\sum_{n=1}^\infty \dfrac{1}{n 2^n}$$ converges quickly, and has nice bounds: $$ \log(1/2) \ge - \sum_{n=1}^{N-1} \dfrac{1}{n 2^n} - \sum_{n=N}^\infty \dfrac{1}{N 2^n} = - \sum_{n=1}^{N-1} \dfrac{1}{n 2^n} - \dfrac{1}{N 2^{N-1}}$$ EDIT: Another way to ...


5

if $a,b,c>0$ then we get $$\frac{2(a^3+b^3+c^3)}{abc}+3\geq \frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}$$ this is equivalent to $$\frac{2(a^3+b^3+c^3)}{abc}\geq \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}$$ and this $$2(a^3+b^3+c^3)\geq bc(b+c)+ac(a+c)+ab(a+b)$$ and now note that $$a^3+b^3=(a+b)(a^2-ab+b^2)\geq ab(a+b)$$


4

Alright, so let's begin by thinking about what this shape looks like. Lets say that we set $z$ to some constant value and see what the cross-section looks like. If $-1\le z\le 1$, then this clearly is empty. If $\left|z\right|>1$, then we get the open disk $x^2+y^2<z^2-1$ (interior of a circle). Now, lets look at those questions: a) Convex: No. ...


4

Note that $1 = (x+y+z)/2$. The inequality becomes $$ \frac{xyz}{(y+z-x)(x+z-y)(x+y-z)} \geq 1$$ which is equivalent to $$ xyz \geq (y+z-x)(x+z-y)(x+y-z)\ \ \ (1)$$ To see why this is true, apply AM-GM inequality three times: $$ x = \frac{x+y-z+x+z-y}{2} \geq \sqrt{(x+y-z)(z+x-y)}$$ and the other two for $y,z$. Now multiply all three and you'll get ...


4

$\le$ is a total order on $\mathbb{R}$. Once $x\le y$, then either $x<y$ or $x=y$. Once $y\le x$, then either $y<x$ or $x=y$. By the trichotomy principle, exactly one of $x<y$, $x=y$, or $x>y$ must hold. Combining these three ideas we conclude $x=y$.


3

Supposing that $3^k\gt k^3$, you need to prove $3^{k+1}\gt (k+1)^3$. For inductive step : $$\begin{align}3^{k+1}&\gt 3k^3\\&=k^3+k^3+k^3\\&\gt k^3+3k^2+3^2k\\&=k^3+3k^2+3k+6k\\&\gt k^3+3k^2+3k+1\\&=(k+1)^3.\end{align}$$


3

Hint: the LHS is an up parabola that has roots at $\pm 2013$. The RHS is also a down parabola with roots at $\pm 2013$. Which values of $y$ satisfy the inequality?


3

So if we accept the Cauchy-Schwarz inequality (a proof can be found here): $$|\langle A, B\rangle| \leq \|A\|\|B\| \iff -\|A\|\|B\|\leq \langle A, B\rangle \leq \|A\|\|B\|$$ Therefore dividing through by $\|A\|\|B\|$, we get: $$-1\leq \frac{\langle A, B\rangle}{\|A\|\|B\|}\leq 1$$ I hope this helps!


3

You arrived at a point where the inequality is equivalent to $$ (a_1b_1+a_2b_2+a_3b_3)^2 \leq (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)$$ You can see that this is true by taking a look at Lagrange's identity. Another way to see this is to look at the discriminant of the following second degree polynomial: $$ p(x)= (a_1x+b_1)^2+(a_2x+b_2)^2+(a_3x+b_3)^2 $$ ...


3

$$x_ny_n-xy=x(y_n-y)+y_n(x_n-x)$$ and then use triangle inequality (norm properties).


3

For $0 <|z| < 1$, $$\sum_{n = 2}^\infty \frac{|z|^n}{n!} \le \sum_{n = 2}^\infty \frac{|z|}{n!} = (e - 2)|z| < \frac{3}{4}|z|.$$


3

Maybe you want to prove: $$ \sum_{k=\color{red}{1}}^{n}\frac{1}{1+\frac{k}{n}}\leq n\log 2\tag{1}$$ that follows from: $$ \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}\leq\int_{0}^{1}\frac{dx}{1+x}=\log 2 \tag{2}$$ since $f(x)=\frac{1}{1+x}$ is a Riemann-integrable, decreasing function over $[0,1]$. If you do not like the integral approach, you may also ...


3

This is not a precise question, yet, something can be said. It is usually resumed in the phrase "inequalities can be integrated but not differentiated". Which is not surprising since integration is essentially the operation of summing, and that preserves inequalities. Differentiation, on the other hand, is more like subtracting, and that does not preserve ...


3

You're almost there. Just find a lower bound for $\frac{|\sin(t)|}{t}$. Since the bounds don't change too much, a first attempt would be $\int_{kπ}^{(k+1)π} \frac{|\sin(t)|}{t}\ dt \ge \int_{kπ}^{(k+1)π} \frac{|\sin(t)|}{(k+1)π}\ dt$. You should be able to continue from here.


2

The upper bound is easy:$$\int_0^{\frac{\pi}{2}}\frac{1+\cos x}{2+\sin x}\mathrm{d}x\leq \int_2^{\frac{\pi}{2}}\frac{1}{2}(1+\cos x)\mathrm{d}x=\pi/4+1/2.$$ For the lower bound, $$\int_0^{\frac{\pi}{2}}\frac{1+\cos x}{2+\sin x}\mathrm{d}x=\int_0^{\frac{\pi}{2}}\frac{1}{2+\sin x}\mathrm{d}x+\int_0^{\frac{\pi}{2}}\frac{\cos x}{2+\sin x}\mathrm{d}x.$$ The ...


2

Needless, because from $k=365$ on, $p_k=0$, so the inequality remains true.


2

Note that $$|\langle A,B\rangle|\leq ||A||\, ||B|| \iff -||A||\, ||B|| \le \langle A,B\rangle\leq ||A||\, ||B|| $$


2

Holder inequality is an attempt to generalize Cauchy-Schwarz inequality to other Lebesgue exponent (other $L^p$ norm). In fact, for inequality of the form $$ \|fg\|_1 \leq \|f\|_p \|g\|_q$$ to be true, we must have $1/p+1/q=1$. To see this, we use scaling argument, which is also useful to verify the correct exponent for other type of inequalities (Sobolev ...


2

This is just an application of Jensen's inequality. As $f(x) = x^2$ is a convex function, we can apply it here. Reference : Jensen's inequality


2

Another way to look at it is to use Titu's Lemma which is also known as Cauchy-Schwarz inequality in Engel form. $$\sum_{i=1}^n\frac{\alpha_i^2}{1}\geq \frac{(\sum_{i=1}^n\alpha_i)^2}{\sum_{i=1}^n1}=\frac{(\alpha_1+\alpha_2+\ldots+\alpha_n)^2}{n}\\ \implies n({\alpha_1^2 + … + \alpha_n^2})\geq (\alpha_1 + … + \alpha_n)^2\\ \implies (\alpha_1 + … + ...


2

Since $\log 2=2\operatorname{arctanh}\frac{1}{3}$, we may use the series: $$ \log 2 = \sum_{n\geq 0}\frac{2}{(2n+1)\,3^{2n+1}} $$ and the bound: $$ \sum_{n\geq 4}\frac{2}{(2n+1)\,3^{2n+1}}\leq\sum_{n\geq 4}\frac{2}{3^{2n+3}}=\frac{1}{157464} $$ to prove that (we just need the terms of the previous series till $n=4$): $$ \log 2\approx \frac{4297606}{6200145} ...


2

Here is the meat of the induction step: \begin{align} 3^{k+1} &= 3^k\cdot 3\tag{by definition}\\[0.5em] &\geq (2^k+1)\cdot 3\tag{by ind. hyp.}\\[0.5em] &=3\cdot 2^k+3\tag{expand}\\[0.5em] &> 2\cdot 2^k+1\\[0.5em] &= 2^{k+1}+1. \end{align}


2

With a log table, you could do it by hand : Compute $\ln(B)=17 \ln(10)$ Then you just have to sum the log of the numbers : $$\ln(n!) = \sum_{k=2}^n \ln(k)$$


2

Your argument is fine and quite clearly presented. You can shorten the presentation considerably, though, by doing something like this: For $n\ge 2$ let $a_n=\sum_{k=n}^{n^2}\frac1k$. Since $a_2=\frac12+\frac13+\frac14>1$, it suffices to show that $a_{n+1}\ge a_n$ for $n\ge 2$. Since $n^2+2n+1<2n^2+n$ for $n\ge 2$, we have ...



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