# Tag Info

16

Hint: $$n!\le n^n=2^{n\log_2 n}$$ Spoiler:

10

You can do it with the AM-GM inequality: $$\frac{a^3+a^3+1}{3}\ge \left(a^3\cdot a^3\cdot 1\right)^{\frac{1}{3}}=a^2$$ $$2\cdot\frac{a^3+1+1+1+1+1}{6}\ge 2\cdot\left(a^3\cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \right)^{\frac{1}{6}}=2\sqrt{a}$$ Adding these inequalities yields the desired result.

7

Let $f(x) = x + 3^x$, then $f'(x) > 0$ for all $x$. Also $f(1) = 4$; however for any $x < 1$, $$f(x) = x + 3^x < 1 + 3 = 4$$ Hence ...

7

For $n=1$ is valid. $$1!\leq 1^1$$ Assume for $n$ is valid: $$n!\leq n^n$$ Multiply by $n+1$ both sides $$(n+1)!\leq n^n(n+1)\leq(n+1)^n(n+1)=(n+1)^{n+1}$$ The last step is because if $n\leq n+1$ then $n^n\leq (n+1)^n$

6

Hint: Prove that $\ln \, x^x \geq \ln x!$ (for $x\geq 0.$) Edit Since $\ln x \geq \ln i,$ for $i \leq x,$ we have $$\ln x+\ln x +\cdots \text{ (x times total)}+\ldots \geq \ln 1 +\cdots \ln x,$$ or $$x \ln x \geq \sum_{i=1}^x \ln i=\ln x!.$$

6

Since,$$\frac{x^5-x^2}{x^5+y^2+z^2} - \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{x^2(x^3-1)^2(y^2+z^2)}{x^3(x^2+y^2+z^2)(x^5+y^2+z^2)} \ge 0$$ Hence, it suffices to prove $\displaystyle \sum\limits_{cyc} \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} \ge 0$ $$\sum\limits_{cyc} \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{\sum\limits_{cyc} \left(x^2 - ... 6 let since$$\begin{cases}a^2-2a\cdot b\cos{\dfrac{2\pi}{3}}+b^2=9\\ b^2-2bc\cos{\dfrac{2\pi}{3}}+c^2=16\\ c^2-2ca\cos{\dfrac{2\pi}{3}}+a^2=25 \end{cases} $$then we let$$PA=a,PB=b,PC=c,\angle APB=\angle BPC=\angle APC=\dfrac{2\pi}{3}$$By cosine$$|AB|^2=25,|BC|^2=16,|AC|^2=9$$so \angle C=\dfrac{\pi}{2}, then we have ... 6 Solution 1: Only use Cauchy-Schwarz inequality :$$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})\ge(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2$$let$$a_{1}=x^{\frac{3}{2}},a_{2}=y^{\frac{1}{2}},a_{3}=1b_{1}=x^{\frac{1}{2}},b_{2}=y^{\frac{3}{2}},b_{3}=1$$so we have$$(x^3+y+1)(x+y^3+1)\ge (x^2+y^2+1)^2$$so$$\sqrt{(x^3+y+1)(y^3+x+1)}\ge x^2+y^2+1$$... 6 I think the hidden agenda here is to prove the statement by induction. If this is true, the inductive hypothesis given in the OP is incorrect, and instead should be$$1+\frac{1}{2}+\cdots+\frac{1}{n-1}>\ln \color{red}{(n-1)}$$To build a proper induction, you begin with the above, add \frac{1}{n} to both sides, then try to prove that$$\ln (n-1) + ...

6

Use Cauchy-Schwarz Inequality: $$(a+b+b+c+c+d+d+a)\left(\frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d}\right) \ge (a+b+c+d)^2$$ Alternative approach: $$\sum\limits_{cyc} \frac{a^2}{a+b} = \sum\limits_{cyc} \left(\frac{a^2}{a+b} - (a-b)\right) = \sum\limits_{cyc} \frac{b^2}{a+b}$$ Therefore, $$\sum\limits_{cyc} \frac{a^2}{a+b} = ... 5 Since A is positive definition, for any t>0 we have$$ A(x+ty)\cdot(x+ty)\geq0. $$That is$$ Ax\cdot x+2tAx\cdot y+t^2Ay\cdot y\geq0,\text{ for all }t\in\mathbb R. $$So the discriminant of above is always non-positive and hence$$ (Ax\cdot y)^2-(Ax\cdot x)\cdot(Ay\cdot y)\leq0. $$That is$$ Ax\cdot y\leq\sqrt{Ax\cdot x}\cdot\sqrt{Ay\cdot y}, $$as ... 4 We know that$$(1+x_1)\cdots(1+x_n) \geq 1+x_1+\ldots +x_n$$If we now multiply the inequality above by (1+x_{n+1}) \geq 0 we obtain$$(1+x_1)\cdots(1+x_n)(1+x_{n+1}) \geq (1+x_1+\ldots +x_n)(1+x_{n+1})\\= (1+x_1+\ldots +x_n) + x_{n+1} + x_{n+1}(x_1+\ldots +x_n) \geq 1 + x_1 + \ldots + x_{n+1}$$where the last inequality follows from the fact that x_i ... 4 Substitute a=b^2, then a^3-a^2-2\sqrt a+2=b^6-b^4-2b+2, which has 1 as a root, so we can factor it out:$$b^6-b^4-2b+2=(b-1)(b^5+b^4-2)$$The polynomial b^5+b^4-2 has still one as a root, so factor more:$$b^5+b^4-2=(b-1)(b^4+2b^3+2b^2+2b+2)So a^3-a^2-2=(b-1)^2(b^4+2b^3+2b^2+2b+2), which is nonnegative for b\ge 0. 4 Why didn't it lead you anywhere? \begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{\frac{2^{(n+1)^{1.001}}}{(n+1)!}}{\frac{2^{n^{1.001}}}{n!}} \\ &= \frac{2^{(n+1)^{1.001}}}{(n+1)2^{n^{1.001}}} \end{aligned} Take the log of both sides. \begin{aligned} \log\left(\frac{a_{n+1}}{a_n}\right) &= ... 4 If ab+bc+ca < \dfrac{-1}{2} \Rightarrow 2(ab+bc+ca) + 1 < 0 \Rightarrow 2(ab+bc+ca) + a^2+b^2+c^2 < 0 \Rightarrow (a+b+c)^2 < 0. Contradiction. 4 Prove that:\log_{10}999^{999}+\log_{10}2<\log_{10}1000^{999}=2997$$In other word:$$\log_{10}2<\log_{10}1000^{999}-\log_{10}999^{999}=\log_{10}\left(\frac{1000}{999}\right)^{999}$$so:$$2<\left(\frac{1000}{999}\right)^{999}=\left(1+\frac{1}{999}\right)^{999}$$It's true by Bernoulli's inequality. Next we should prove 3 \cdot ... 3 0\leq(a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc)\implies ab+ac+bc\geq-\dfrac{1}{2}. 3 I really wouldn't call the argument you mention "heuristic." Simply note that$$\dfrac{n^n}{n!}=\frac{n}{n} \frac{n}{n-1} \frac{n}{n-2}\cdots \frac{n}{2} \frac{n}{1}\geq 1$$This is because every term in the product is larger than or equal to 1. 3 We have,$$\sum_{\mathbf{x}\in\mathcal{S}}\exp\left[f(n)\sum_{i=1}^nx_i\right] = \sum_{\mathbf{x}\in\mathcal{S}} \prod_{(x_i)_{i=1}^{n} = \mathbf{x}} e^{f(n)x_i}=\left(e^{f(n)}+e^{-f(n)}\right)^n$$The identity is a consequence of: \displaystyle \left(t^{1}+t^{-1}\right)^{n} = \sum\limits_{\substack{1 \le i \le n\\ \epsilon_i = \pm1}} ... 3 We have that there are n\choose r vectors in S with exactly r negative components, and that the sum of the components of any one of these vectors is n-2r, so your sum is$$A_{f(n)}(n)=\frac{1}{2^n}\sum_{\mathbf{x}\in\mathcal{S}}\exp\left[f(n)\sum_{i=1}^nx_i\right] = \frac{1}{2^n}\sum_{r=0}^n {n \choose r} \exp(f(n)(n-2r))\le ...

3

There are two cases. Either $a>0$ or $a<0$. ($a$ must be nonzero since the problem refers to $1/a$) case 1) $a>0\Rightarrow a<\frac{1}{a} \Rightarrow a^2 < 1\Rightarrow 0<a<1$ Consider what happens with the addition of $b$ into the mix. Since $0<a<\frac{1}{a}<b<\frac{1}{b}$, you have that $0<a<b$ and $b$ is ...

3

We can do better if we incorporate the offending inequality into the mix: $$6(x+y+z)\le 2(x+2y+z)+(x+y+2z)+(3x+y+2z)\le2\cdot60+70+98=288\\\implies x+y+z\le 48$$ Setting equality in these three inequalities we used to compute this bound yields $(x,y,z)=(14, 12, 22)$, which satisfies the remaining inequalities and confirms $48$ is the answer.

3

By the triangle inequality we have $$|\lambda_1 a_1 + ... +\lambda_n a_n|\leq |\lambda_1||a_1|+...+|\lambda_n||a_n|$$ Since each $|a_i|<1$ and each $\lambda_i$ is non-negative, we have $|\lambda_i|= \lambda_i$. Then $$|\lambda_1||a_1|+...+|\lambda_n||a_n| < \sum_i \lambda_i =1$$ Since you tagged this as complex analysis, I assume you meant for the ...

3

I would like to suggest "different" approach. First off as noted in one of the comments $\lambda_i$ must be real since $\mathbb{C}$ is not an ordered field. On the another hand $\mathbb{C}$ is a vector space over the field of real numbers. So I am going to translate your original statement into a geometric interpretation. Imagine that you pick $n$ points in ...

3

Usually it's referred to as the triangle inequality.

3

There are many ways to prove this, and the preferred method will probably depend on your axioms and previously proven results. One way is as follows: Consider $y=1+\epsilon$ for $\epsilon>0$. By the binomial theorem (the terms we are dropping are all positive), $y^n=(1+\epsilon)^n\geq 1+n\epsilon$. Setting $N=\frac{M}{\epsilon}$ then yields the result ...

3

By C-S $\left(\sum\limits_{cyc}\sqrt{\frac{y^2+z^2}{2(x^2+y^2)(x^2+z^2)}}\right)^2\leq\sum_(y^2+z^2)\sum\limits_{cyc}\frac{1}{2(x^2+y^2)(x^2+z^2)}=\frac{2(x^2+y^2+z^2)^2}{\prod\limits_{cyc}(x^2+y^2)}$. Thus, it remains to prove that $\left(\sum\limits_{cyc}(x^5+x^3y^2+x^3z^2+x^2y^2z)\right)^2\geq2(x^2+y^2+z^2)^2(x^2+y^2)(x^2+z^2)(y^2+z^2)$. Let $x+y+z=3u$, ...

3

That's a private case of AM-GM: $$(\sqrt a-\sqrt b)^2>0\Leftrightarrow a+b-2\sqrt {ab}>0\Leftrightarrow\frac{a+b}{2}>\sqrt{ab}$$

3

The power series of the exponential function is defined on $\Bbb R$ so we can differentiate it term by term on $\Bbb R$ and we get $$\exp'(x)=\exp(x)$$ Moreover, we see easily that $\exp(x)>0$ for $x\ge0$ and using the Cauchy product we get $$\exp(x)\exp(y)=\exp(x+y),\quad \forall (x,y)\in\Bbb R^2$$ hence \exp(-x)\exp(x)=\exp(0)=1\implies ...

3

Another approach using $4ab\le(a+b)^2$: \begin{align*}\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}&=\left(a-\frac{ab}{a+b}\right)+\left(b-\frac{bc}{b+c}\right)+\left(c-\frac{cd}{c+d}\right)+\left(d-\frac{da}{d+a}\right)\\ &=1-\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{cd}{c+d}+\frac{da}{d+a}\right)\\ ...

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