星期四��上21点，将会播出一集女����的����。这个����将会������都市��拉利�����������和����。女��们��供��在����上，星期四��上21点，将会播出一集女����的����。这个����将会������都市��拉利�����������和����。女��们��供��在����上，其中也包����称为“美之女��”的����。��尔·������从���������生还后，本应该享��这场��会。然而，直到他收到一��来自��位����������的信，他��意��到这次女����可能会有些不同��常。信中提到，��尔��和他���自约会。在����都市的��个角��，少女��注一��地��出��定，这使得��尔和整个����都市都��入了������之中。与此同时，号称“最��”的“���������们”也开始行动了。这将是少年的���步，少女的期�� - 【��族之物语】。<|endoftext|> \begin{tikzpicture}
% Define nodes
\node[obs] (x) {$\mathbf{x}$};
\node[latent, above=of x, xshift=-0.5cm] (z) {$\mathbf{z}$};
\node[latent, above=of x, xshift=0.5cm] (theta) {$\boldsymbol{\theta}$};
\node[latent, below=of x, xshift=0.75cm] (w) {$\mathbf{w}$};
% Connect the nodes
\edge {z} {x}
\edge {theta} {x}
\edge {w} {x}
% Plates
\plate {plate1} {(x)(z)(theta)} {$N$};
\plate {plate2} {(x)(w)} {$M$};
\end{tikzpicture}<|endoftext|>Varje tal som är summan av två perfekta kvadrater är delbar med fyra.
<|endoftext|>## Problem 1
Evaluate the following numerical expressions.
a) $3 4$
b) $7-2$
c) $5-8$
d) $12 10$
a) $3 4 = 7$
b) $7-2 = 5$
c) $5-8 = -3$
d) $12 10 = 22$<|endoftext|>Algebraic<|endoftext|>#define ID 1
#define NAME "John Doe"
#define AGE 30
#define PI 3.14159265359<|endoftext|>## Problem 1
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$
\frac{a}{a^2 1} \frac{b}{b^2 1} \frac{c}{c^2 1} \leq \frac{9}{10}.
$$
Solution:
We begin by noting that for any positive real number $x$, we have
$$
x^2 1 \geq 2x,
$$
since this inequality is equivalent to
$$
x^2-2x 1=(x-1)^2 \geq 0,
$$
which is clearly true. Hence, for any positive real numbers $a, b, c$, we have
$$
\frac{a}{a^2 1} \leq \frac{a}{2a} = \frac{1}{2}.
$$
Similarly, we have $\frac{b}{b^2 1} \leq \frac{1}{2}$ and $\frac{c}{c^2 1} \leq \frac{1}{2}$. Summing these three inequalities, we obtain
$$
\frac{a}{a^2 1} \frac{b}{b^2 1} \frac{c}{c^2 1} \leq \frac{1}{2} \frac{1}{2} \frac{1}{2} = \frac{3}{2}.
$$
However, since $a b c=1$, we have $a, b, c < 1$, and hence $a^2, b^2, c^2 < 1$. This implies that $a^2 1, b^2 1, c^2 1 < 2$, and therefore, $\frac{1}{a^2 1}, \frac{1}{b^2 1}, \frac{1}{c^2 1} > \frac{1}{2}$. Multiplying these three inequalities, we obtain
$$
\frac{1}{(a^2 1)(b^2 1)(c^2 1)} > \frac{1}{2^3} = \frac{1}{8}.
$$
Raising both sides of this inequality to the $3$rd power, we obtain
$$
\frac{1}{(a^2 1)^3(b^2 1)^3(c^2 1)^3} > \frac{1}{8^3} = \frac{1}{512}.
$$
Multiplying both sides of this inequality by $a^3b^3c^3$, we obtain
$$
\frac{a^3b^3c^3}{(a^2 1)^3(b^2 1)^3(c^2 1)^3} > \frac{a^3b^3c^3}{512}.
$$
However, by the AM-GM inequality, we have
$$
\frac{a^3b^3c^3}{512} \leq \left(\frac{a b c}{3}\right)^9 = \frac{1}{3^9} = \frac{1}{19683}.
$$
Combining this inequality with the previous one, we obtain
$$
\frac{a^3b^3c^3}{(a^2 1)^3(b^2 1)^3(c^2 1)^3} > \frac{1}{19683}.
$$
Finally, taking reciprocals on both sides of this inequality, we obtain
$$
(a^2 1)^3(b^2 1)^3(c^2 1)^3 > 19683,
$$
which implies that
$$
(a^2 1)(b^2 1)(c^2 1) > 27.
$$
Hence, we have
$$
\frac{1}{a^2 1} \cdot \frac{1}{b^2 1} \cdot \frac{1}{c^2 1} < \frac{1}{27}.
$$
Multiplying both sides of this inequality by $a^2b^2c^2$, we obtain
$$
\frac{a^2b^2c^2}{(a^2 1)(b^2 1)(c^2 1)} < \frac{a^2b^2c^2}{27}.
$$
However, by the AM-GM inequality, we have
$$
\frac{a^2b^2c^2}{27} \leq \left(\frac{a b c}{3}\right)^6 = \frac{1}{3^6} = \frac{1}{729}.
$$
Combining this inequality with the previous one, we obtain
$$
\frac{a^2b^2c^2}{(a^2 1)(b^2 1)(c^2 1)} < \frac{1}{729},
$$
and hence,
$$
\frac{a^2}{a^2 1} \cdot \frac{b^2}{b^2 1} \cdot \frac{c^2}{c^2 1} < \frac{1}{729}.
$$
Multiplying both sides of this inequality by $a^2 b^2 c^2$, we obtain
$$
\frac{a^2}{a^2 1} \cdot \frac{b^2}{b^2 1} \cdot \frac{c^2}{c^2 1} \cdot (a^2 b^2 c^2) < \frac{a^2 b^2 c^2}{729}.
$$
However, by the Cauchy-Schwarz inequality, we have
$$
(a^2 b^2 c^2)\left(\frac{a^2}{a^2 1} \frac{b^2}{b^2 1} \frac{c^2}{c^2 1}\right) \geq (a b c)^2 = 1,
$$
and hence,
$$
\frac{a^2}{a^2 1} \frac{b^2}{b^2 1} \frac{c^2}{c^2 1} \geq \frac{1}{a^2 b^2 c^2} > 1.
$$
Combining this inequality with the previous one, we obtain
$$
\frac{a^2}{a^2 1} \cdot \frac{b^2}{b^2 1} \cdot \frac{c^2}{c^2 1} \cdot (a^2 b^2 c^2) < \frac{1}{729} \cdot \frac{a^2 b^2 c^2}{a^2 b^2 c^2} = \frac{1}{729}.
$$
Finally, multiplying both sides of this inequality by $9$, we obtain
$$
\frac{9a^2}{a^2 1} \cdot \frac{9b^2}{b^2 1} \cdot \frac{9c^2}{c^2 1} < \frac{9}{729},
$$
which simplifies to
$$
\frac{a^2}{a^2 1} \cdot \frac{b^2}{b^2 1} \cdot \frac{c^2}{c^2 1} < \frac{1}{81}.
$$
Combining this inequality with the inequality we obtained earlier, we obtain
$$
\frac{a^2}{a^2 1} \cdot \frac{b^2}{b^2 1} \cdot \frac{c^2}{c^2 1} < \frac{1}{81} < \frac{1}{27} < \frac{1}{8} < \frac{1}{2} < \frac{3}{2}.
$$
Therefore, we have
$$
\frac{a^2}{a^2 1} \frac{b^2}{b^2 1} \frac{c^2}{c^2 1} < \frac{3}{2}.
$$
Combining this inequality with the inequalities we obtained at the beginning of the solution, we obtain
$$
\frac{a}{a^2 1} \frac{b}{b^2 1} \frac{c}{c^2 1} < \frac{1}{2} \frac{1}{2} \frac{1}{2} = \frac{3}{2} < \frac{9}{10}.
$$
Hence, we have
$$
\frac{a}{a^2 1} \frac{b}{b^2 1} \frac{c}{c^2 1} < \frac{9}{10},
$$
as desired. $\square$<|endoftext|># Language: Python 3 Notebook
# Language: Python
#
1. Define详情
