# Tag Info

18

You are essentially asking whether $$\frac 56 \ \ \ \ \text{ or }\ \ \ \ \left(\frac{35}{36}\right)^6$$ are bigger. Now by the Bernoulli's inequality, $$\left(\frac{35}{36}\right)^6 = \left(1- \frac{1}{36}\right)^6 > 1 - 6\frac{1}{36} = \frac 56.$$

11

Using trapezoidal integration approximation to $\int_0^1x^ndx$ with step $\frac1n$ $$\Bigl(\frac 1n\Bigr)^n + \Bigl(\frac 2n\Bigr)^n + \cdots + \Bigl(\frac nn\Bigr)^n>n\int_0^1x^ndx+\frac12\Bigl(\frac nn\Bigr)^n=\frac{3n+1}{2n+2}$$

9

Let $f(x)=x^{r+1}+\log(r+1)-\sum_{i=1}^r{x^i/i}$. We need to show that $\min_{x>0}f(x)>0$. Differentiate and set to $0$: $$f'(x)=(r+1)x^r-\sum_{i=0}^{r-1}{x^i}=0$$ Clearly $f'(t)>0$ for $t\geq 1$, so solution(s) for $f'(x)=0$ lies in $(0,1)$ (at least one exists since $f'(0)<0$ and $f'$is continuous). Use $x^r=\sum_{i=0}^{r-1}{x^i}/(r+1)$, ...

7

Note that for any $x$, we have $$(e^{x^2} + x) - e^x = \\ \frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \left(\frac{1}{2!} - \frac{1}{4!}\right)x^4 - \frac 1{5!}x^5 + \left( \frac{1}{3!} - \frac{1}{6!} \right)x^6 - \cdots \geq \\ \frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \frac {1}{4!}x^4 - \frac 1{5!}x^5 + \frac{1}{6!}x^6 - \cdots =\\ e^{-x} + x - 1$$ Consider the ...

5

Hint. The inequality basically states $$\left|\Im(z)\right| \leq \left|z\right| = \sqrt{\Re(z)^2+\Im(z)^2} \qquad z \in \Bbb C$$ For an intuitive understanding just think of the complex plane.

4

You could also subtract. $$\frac1n - \frac{1}{n+1} = \frac{1}{n(n+1)} > 0.$$ Since the difference is positive, $\frac1n$ must be the larger one.

4

Let $x = \frac{a}b+\frac{b}a \ge 2$ (by AM-GM). Then $$x^3=\frac{a^3}{b^3}+\frac{b^3}{a^3} + 3x$$ So we want to show, for $x \ge 2$, $$3(x^3-3x) \ge x+4 \iff (x-2)(3x^2+6x+2) \ge 0$$ which is obvious.

4

By AM-GM: $$\frac{a^2}{a+b}+\frac{a+b}{4}\ge a$$ and $$\frac{b^2}{b+c}+\frac{b+c}{4}\ge b.$$ So $$\frac{a^2}{a+b}+\frac{b^2}{b+c}\ge a+b-\frac{a+b}4-\frac{b+c}4=\frac{3a+2b-c}4.$$

3

Raise both to the power of $3\over2$. Then we will be comparing $\left(\frac{5}{6}\right)^6\;\text{ to }\;\left(\frac{35}{36}\right)^{36}$, and we might have heard a thing or two about the sequence $\left(\frac{n-1}{n}\right)^n$, which increases and converges to... well, that's not really important here, just note that it increases.

3

$$\log_x\left(y^z\right)=z\cdot\log_xz$$ and $$\log_xz=\dfrac{\log z}{\log x}$$ Now $$z\cdot\dfrac{\log z}{\log x}>0$$ as $x,y,z>1$ Using AM-GM inequality $$\dfrac{\sum z\dfrac{\log y}{\log x}}3\ge\sqrt[3]{xyz\prod\dfrac{\log y}{\log x}}=\sqrt[3]{xyz}>1$$ as $x,y,z>1$

3

The LHS is non-negative and $x_1^2+x_2^2\leq 1$, hence the inequality is non-trivial only if $y_1^2+y_2^2\leq 1$. So we may assume that $X=(x_1,x_2)$ and $Y=(y_1,y_2)$ are two points inside the unit circle and the angle between them is $\theta$, then prove: $$\left(\|X\|\|Y\|\cos\theta-1\right)^2\geq (1-\|X\|^2)(1-\|Y\|^2) \tag{1}$$ that is equivalent to: ...

3

it is just $$\left| \frac {1}{x}-\frac{1}{z} \right|=\left|\frac {1}{x}-\frac {1}{y}+\frac{1}{y}-\frac{1}{z}\right| \leq \left|\frac {1}{x}-\frac{1}{y}\right|+\left|\frac {1}{y}-\frac{1}{z}\right|$$ using the Module properties

3

Hints: Show that the following inequalities follow from the given assumptions: $$0\leq(1+a)(1+b),\quad 0\leq(1-a)(1-b).$$ Then expand them.

3

Since the inequality is homogeneous, we may assume without loss of generality that $y=1$ and $x\in(0,1]$, then prove: $$\forall x\in(0,1],\qquad (1+x)^p > 1+x^p.\tag{1}$$ On the other hand, if we set $f(x)=(1+x)^p-x^p$, it is straightforward to check that $f'(x) = (p-1)\cdot\left((1+x)^{p-1}-x^{p-1}\right)>0$, hence $f$ is increasing on $(0,1]$. Since ...

3

One other way is to consider $f(x)=e^{x^2}+x-e^x$, we have $f'(x)=2xe^{x^2}+1-e^x$ and $f''(x)=(4 x^2+2) e^{x^2}-e^x$, So $$f''(x)=e^x\left(e^{x^2-x}(4x^2+2)-1\right)\ge e^x\left(e^{-1/4}\times 2-1\right)>0$$ Since $x^2-x\ge-1/4$. So, $f$ is convex, and because $f'(0)=0$ we conclude that $f$ attains its minimum at $x=0$ which is $0$. The desired ...

2

You can use that $a>b>0$ implies $\frac{1}{b}>\frac{1}{a}>0$. You have $n+1>n$ so $\frac{1}{n}>\frac{1}{n+1}.$

2

No. In general, for $a,b,c,d\in \Bbb N$ with $\frac{a}{b} < \frac{c}{d}$, $$\frac{a}{b} <\frac{a+c}{b+d}<\frac{c}{d}.$$ Proof: The left inequality is equivalent to $$a(b+d) < b(a+c) \iff ad < bc \iff \frac{a}{b}<\frac{c}{d}.$$ Likewise for the right inequality.

2

There cannot be such a bound, since the constraint is compatible with arbitrarily high variance and hence, by the central limit theorem, high deviation probabilities for arbitrarily high $n$. The variance of $X$ with $P(X=0)=1-\epsilon$ and $P(X=M/\epsilon)=\epsilon$, with expectation $M$, is (1-\epsilon)M^2+\epsilon ... 2 For a,b\in\mathbb R, we have |a+bi|^2 = a^2 + b^2 \ge b^2, so |a+bi| \ge |b|. 2 Hint: Since x^2=|x|^2 let's make the sustitution y=|x| and solve y^2-y-2<0 and consider these values of y\ge 0. Since y^2-y-2=\left(y-\frac{1}{2}\right)^2-\frac{9}{4} we have \begin{align} y^2-y-2<0 \quad&\iff & \left(|x|-\frac{1}{2}\right)^2 &< \frac{9}{4}\\ &\iff & -\frac{3}{2}<|x|-\frac{1}{2}&<\frac{3}{2} ... 2 From AM-GM, we have: x^4+y^4+z^2=x^4+y^4+\frac12 z^2+\frac12 z^2\ge4\left(x^4y^4\frac12 z^2\frac12 z^2\right)^{\frac14}=\sqrt8xyz $$A simple proof of this inequality for 4 variables: Firstly, we have for a,b\ge0:$$ \frac{a+b}{2}\ge\sqrt{ab}\iff \left(\sqrt a-\sqrt b\right)^2\ge0 $$Therefore:$$ ...

2

A better idea is to solve for $x$: $$(7+2a)x < -21-6 = -27.$$ If $7+2a>0$, dividing gives an inequality of the form $x<\cdots$, which we don't want. If $7+2a<0$, dividing by it reverses the sense of the inequality, so $$x > \frac{-27}{7+2a}.$$ You're looking for this to be satisfied by all $x>3$, so therefore the right-hand side must be ...

2

Obviously for any $1>t>0$ and $p>1$ we have $t>t^p$ thus $$\frac{x}{x+y} > \left( \frac{x}{x+y} \right)^p \hbox{ and } \frac{y}{x+y} > \left( \frac{y}{x+y} \right)^p.$$ Summing up and multiplying by $(x+y)^p$ we get $(x+y)^p > x^p+y^p$.

2

Since $0 < \dfrac{x}{x+y} < 1, 0 < \dfrac{y}{x+y} < 1 \Rightarrow \left(\dfrac{x}{x+y}\right)^p < \dfrac{x}{x+y}, \left(\dfrac{y}{x+y}\right)^p < \dfrac{y}{x+y}$. Adding these inequalities, the answer follows.

2

A re-worked answer. We have: $$e^n = \sum_{k=0}^{n}\frac{n^k}{k!}+\frac{n^{n+1}}{n!}\int_{0}^{1}\left(e^t(1-t)\right)^n\,dt\tag{1}$$ but we also have: $$\forall x\in[0,1],\qquad (1-x^3)e^{-x^2/2} \leq e^{x}(1-x) \leq e^{-x^2/2}\tag{2}$$ hence: $$\int_{0}^{1}\left(e^t(1-t)\right)^n\,dt \leq \int_{0}^{+\infty}e^{-nx^2/2}\,dx = \sqrt{\frac{\pi}{2n}}\tag{3}$$ ...

2

you need just add all the inequalities, and you would have: $$|ax-b|+|bx-c|+|cx-a|\leq a+b+c$$ Also, from the Module properties: $$|ax-b+bx-c+cx-a|\leq |ax-b|+|bx-c|+|cx-a|$$ $$|(a+b+c)x-(a+b+c)|\leq |ax-b|+|bx-c|+|cx-a|\leq a+b+c$$ Dividing by $(a+b+c)$, we have: $$|x-1|\leq 1$$ which proves that $0 \leq x \leq 2$. Division would not affect the inequality ...

2

With the current version of the problem again let $a^2 = x-1, b^2 = y-1$, then you want to show $$(a^2+1)b + (b^2+1)a \le (a^2+1)(b^2+1)$$ $$\iff (b^2+1-b)a^2-(b^2+1) \cdot a+(b^2+1-b) \ge 0$$ $$\iff a^2-\frac{b^2+1}{b^2+1-b} \cdot a+1 \ge 0$$ $$\iff \left(a-\frac{b^2+1}{2(b^2-b+1)} \right)^2+\frac{(b-1)^2(3b^2-2b+3)}{4(b^2-b+1)^2} \ge 0$$ which is obvious ...

2

You're almost done. If you have $2 \leq x \leq 5$ then isn't it clear that $x^2 > 1$? By the way, I don't think proving the contrapositive made this any easier. Just start by factoring $x^2 \leq 1$ and deduce from there.

2

Not a very elegant solution, I will admit. We want to show that $$\frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}\leq 4$$ which is equivalent to (since the denominator is non-negative by AM-GM) $$(a+b+c)^3-27abc \leq 4(a^3+b^3+c^3) - 12abc$$ or $$a^3 + b^3 + c^3 + 3a^2(b+c) + 3b^2(c+a)+3c^2(a+b) + 6abc - 27abc \leq 4(a^3+b^3+c^3)-12abc$$ which further simplifies ...

2

this inequation is equivalent to $$\frac{(a-b)^2 \left(3 a^4+6 a^3 b+8 a^2 b^2+6 a b^3+3 b^4\right)}{a^3 b^3}\geq 0$$ which is true.

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