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8

We have to prove that: $$ \sum_{k=1}^{2n}\sqrt{2k-1}-\sum_{k=1}^{2n}\sqrt{2k-2} > \sqrt{n} \tag{1}$$ hence it looks like a good idea to apply creative telescoping and approximate: $$\sqrt{2k-1}-\sqrt{2k-2}\geq\frac{\sqrt{k-1/4}-\sqrt{k-5/4}}{\sqrt{2}}-\frac{1}{128\sqrt{2}}\left(\frac{1}{(k-5/4)^{3/2}}-\frac{1}{(k-1/4)^{3/2}}\right)\tag{2} $$ I found $(2)$ ...


7

Hint: $$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}=$$ $$=\frac{a}{(1-a)^2}+\frac{b}{(1-b)^2}+\frac{c}{(1-c)^2}$$ Let $f(t)=\frac t{1-t^2}$ Use Jensen's Inequality


6

It is hardly a lot of algebra. To save on MathJax, I write $\sqrt n$ as $s$. So the square of the RHS is $(s+\frac{1}{2s}-\frac{1}{8s^3})^2=1+s^2+\frac{1}{64s^6}-\frac{1}{8s^4}$. The square of the LHS is just $1+s^2$. So you have to show that $\frac{1}{8s^4}>\frac{1}{64s^6}$ or $8s^2>1$ or $8n>1$ which is true for any positive integer and indeed ...


5

It is known as the Shafer-Fink inequality. Here you may find different proofs and some improvements, too.


5

Note that $$ \sum_{n=0}^\infty \dfrac{1}{n^2 + 1}= \sum_{n=0}^\infty\int_{n}^{n+1}\frac{1}{n^2+1}dx>\sum_{n=0}^\infty\int_{n}^{n+1}\frac{1}{x^2+1}dx=\int_{0}^\infty\frac{1}{x^2+1}dx=\frac{\pi}{2}. $$ Also note $$ \sum_{n=0}^\infty \dfrac{1}{n^2+1}=1+\sum_{n=1}^\infty \dfrac{1}{n^2 + 1}<1+\sum_{n=1}^\infty \dfrac{1}{n^2}=1+\frac{\pi^2}{6}<\frac{3\...


5

Note that: $$a^2c^2 + a^2b^2 \ge 2a^2bc \quad\text{ by AM-GM}$$ Now add all the cyclic inequalities and you'll get the wanted inequality.


4

Let $g(x) = f^2(x) + f'^2(x)$. We want to show that $g$ is bounded by $1$ when $f^2(x) \leq 1$ and $f'^2(x) + f''^2(x) \leq 1$. The extremal value(s) of $g$ are found at points $x_*$ where $g'(x_*) = 2f'(x_*)(f(x_*) + f''(x_*)) = 0$. If $f'(x_*) = 0$ then $g(x_*) = f^2(x_*) \leq 1$ and if $f(x_*) = -f''(x_*)$ we have $g(x_*) = f'^2(x_*) + f''^2(x_*) \leq ...


4

Squaring both sides, $$a^2x^2-2abx+b^2 \leq b^2x^2-2abx+a^2$$ $$(a^2-b^2)x^2 \leq a^2-b^2$$ Since $a^2-b^2>0$, we can divide both sides by $a^2-b^2$ and preserve the sign of inequality. $$x^2 \leq 1$$ $$|x|\leq 1$$ $$-1 \leq x \leq 1$$


4

\begin{align} \left|\bigcup_{v \in D} N(v)\right| &\leq \sum_{v \in D} |N(v)| \\ &= \sum_{v \in D} \deg_G(v) \\ &\leq \sum_{v \in D} \Delta(G) \\ &=|D|\cdot\Delta(G) \end{align} I hope this helps $\ddot\smile$


4

It does not look so terrible to me! We have: $$ \sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}} \tag{1}$$ and: $$ \frac{1}{2\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{1}{2\sqrt{n}\left(\sqrt{n}+\sqrt{n+1}\right)^2}\tag{2}$$ so the inequality boils down to $\sqrt{n}+\sqrt{n+1}>2\...


4

If $\beta \ge 1$, this is a consequence of Jensen's inequality applied to the convex function $t \mapsto t^{\beta}$. If $\beta < 1$, this is false with $x_i = 1$.


4

Note that this amounts to showing that $$ \left(\sum_{i=1}^{N}{x_i}^{\beta}\right)^{1/\beta} \leq \sum_{i=1}^N x_i $$ This fact holds for all $\beta \geq 1$, and it is an instance of the Minkowski inequality.


4

Since $a,b,c>0$, \begin{align} \frac{a \log a + b \log b + c \log c}{a+b+c} \le \log \left(\sum_{cyc}a\times \frac{a}{a+ b+ c}\right) \end{align} by Jensen's inequality on the $\log$. Taking exponent gives the required result.


4

Note that $Y$ is also the number of integral solutions $\left(z_1,z_2,\ldots,z_n\right)$ satisfying the condition $0\leq z_1\leq z_2\leq \ldots\leq z_n$ to the equation $$z_1+z_2+\ldots+z_n=\frac{n(n-1)}{2}$$ (i.e., by setting $z_k:=y_k-k$ for every $k=1,2,\ldots,n$). Hence, $n!\cdot Y$ is at least the number of integral solutions $\left(w_1,w_2,\ldots,w_n\...


3

Let's say $x=\tan \theta$, so that: $$\theta \geq \frac{3\tan \theta}{1+2\sqrt{1+\tan^2 \theta}}$$ $\sqrt{1+\tan^2\theta}=\sec\theta$ because the range of $\arctan x$ always has $\cos \theta > 0$: $$\theta \geq \frac{3\tan \theta}{1+2\sec\theta}$$ Multiply both the numerator and denominator by $\cos \theta$: $$\theta \geq \frac{3\sin\theta}{\cos\theta+2}$$...


3

Note that $A_n$ is also equal to the number of tuples $(y_1,y_2,\ldots,y_n)\in\mathbb{Z}^n$ such that $0<y_1<y_2< \ldots y_n$ and $y_1+\ldots +y_n=n^2$. [Consider the bijection $y_1=x_n,y_2=x_n+x_{n-1}$,$\ldots$,$y_n=x_n+\ldots +x_1$.] The number of such tuples is less than $\dfrac{1}{n!}$ times the number of positive integer solutions to $z_1+\...


3

1) $0 > -x^2 + 2x + 3$ so $0> (-x+3)(x+1)$ So $(-x + 3)(x+1)$ is negative. So one of the terms is positive and the other is negative. So either $-x +3 > 0$ AND $x+1 < 0$ OR $-x + 3 < 0$ AND $x+1 > 0$ If $-x +3 > 0$ AND $x+1 < 0$ then $x < 3$ and $x < -1$. Notice these are redundant statements. If $x < -1$ then OF ...


3

You've already solved it!! You've shown that $\log([x^2+y^2]/4)\leq x+y-2$. You just have to take exponentials on both sides and (because the exponential is an increasing map) you're done. [It might be noteworthy that your demostration is for the case that $x$ and $y$ are not simultaneously zero, but that case is trivially true] Edit: Not an answer, yet. I ...


3

We have: $$ -1+2\sum_{n\geq 0}\frac{1}{n^2+1}=\sum_{n\in\mathbb{Z}}\frac{1}{n^2+1}\color{red}{=}\sum_{n\in\mathbb{Z}}\pi e^{-2\pi|n|}=\pi\coth(\pi) \tag{1}$$ where $\color{red}{=}$ holds by the Poisson summation formula. Since $\coth(\pi)>\pi$ but $\coth(\pi)<\pi+e^{-\pi}$, $$ \frac{\pi+1}{2}<\sum_{n\geq 0}\frac{1}{n^2+1}<\frac{\pi+1+e^{...


3

The general theorem of means (perhaps more popularly known as Young's inequality as the comment to this answer suggests) implies that if $x, y, r, s$ are positive with $r + s = 1$ then $$x^{r}y^{s} \leq rx + sy\tag{1}$$ Putting $x = a^{p}, y = b^{q}, r = 1/p, s = 1/q$ we get $$ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}$$


3

Hint: Set $1/a=x$ etc. and use $(x-y)^2\ge0$ for real $x,y$


3

The case $0<\beta<1$ we have re reversed inequality $$\left(\sum_{i=1}^N x_i\right)^\beta\leq \sum_{i=1}^N x_i^\beta$$ Also, this can actually be used to prove the inequality of the present post. Here are some details (for the case $\beta>1$ choose $p=1$ and $q=\beta$).


3

$$\left| \left| x-2 \right| -3 \right| > 1$$ is the same as $$ \left| x-2 \right| < 2 \text{ or } \left| x-2 \right| >4$$ which further reduces to $$0<x<4 \text{ or } x<-2 \text{ or } x>6$$ I think this will be enough to sketch the graph.


3

Multiply the inequality by $\;(x+2)^2>0\;$ ( obviously, $\;x\neq-2\;$) : $$x(x+2)^2<x+2\iff x^3+4x^2+3x-2<0\iff$$ $$\iff (x+2)(x+1-\sqrt2)(x+1+\sqrt2)<0\iff \color{red}{x<-1-\sqrt 2}\;\;\text{or}\;\color{red}{-2<x<-1+\sqrt2}$$


3

solutions you found are not correct. Correct solutions are $y\le -1$ and $y\ge 2$ but first solution gives no value of $x$ becuase $y$ is always positive. So now $3^x\ge 2$ gives $x\ge log_32$ which is the final answer. Hope this helps !


3

If $a,b$ or $c$ equals zero the inequality is trivial. Hence we may assume: $$ (a,b,c)=\left(e^{A},e^{B},e^{C}\right) $$ without loss of generality, and by taking $f(x)=e^{x}$ the inequality can be written as: $$ \frac{f(A)+f(B)+f(C)}{3}\geq f\left(\frac{A+B+C}{3}\right) $$ that holds as a consequence of the convexity of $f(x)$.


3

Note that the sequence you're interested in can be written as $a_n=\dfrac1{\sqrt{n}}\left(\sqrt{4n-1}+\sum\limits_{k=1}^{2n-1}\sqrt{2k-1}-\sqrt{2k}\right).$ Applying the Stolz-Cesàro theorem, we have \begin{align}\lim_{n\to\infty} a_n&=\lim_{n\to\infty}\frac{\sqrt{4n+3}-\sqrt{4n-1}+\sqrt{4n-1}-\sqrt{4n}+\sqrt{4n+1}-\sqrt{4n+2}}{\sqrt{n+1}-\sqrt{n}} \\ &...


3

You have got yourself into a tangle by being too clever too soon, and squaring. It is better to simplify first. With careful simplification you don't need calculus or anything advanced at all. To pick up a theme of Kenny Lau, put $u=x+\sqrt{x^2-10x+9}$. Then your inequality becomes $u\ge\sqrt{u}$, which is equivalent (following Gerry Myerson's comment) to $...


2

Using the approximation $\sqrt3\approx1.7$, one can do a bit of mental arithmetic and see that $${3\sqrt3-4\over7-2\sqrt3}\approx{1.1\over3.6}\lt1\qquad\text{while}\qquad{3\sqrt3-8\over1-2\sqrt3}\approx{-2.9\over-2.4}={2.9\over2.4}\gt1$$ This only works, of course, because the two numbers are not at all close. It's also not a rigorous proof, although it ...


2

$\require{cancel}$ Generally speaking, if we have two relations $R$ and $S$, then we define $$a\,\cancel {R\,}\,b\iff \lnot(a\,R\,b)\\ a\,{}^R_S\,b\iff (a\,R\, b) \vee(a\,S\,b).\tag{*}\label{moi}$$ There are some exeptions, which I personally try to avoid, like $A\subsetneqq B$. Before I go on, I must point out a crucial inconsistency in the OP. In ...



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