## 2019年10月12日土曜日

### 数学 - Python - 急速・緩慢に変化する関係 - 指数関数・対数関数 - 対数関数の性質 - いくつかの例題および問題の補充 - 不等式、式の変形

1. $\begin{array}{l}\left({\mathrm{log}}_{a}\frac{x}{y}{\mathrm{log}}_{a}\frac{x}{z}\right)-\left({\mathrm{log}}_{a}\sqrt{\frac{y}{z}}·{\mathrm{log}}_{a}\sqrt{\frac{z}{y}}\right)\\ =\left({\mathrm{log}}_{a}x-{\mathrm{log}}_{a}y\right)\left({\mathrm{log}}_{a}x-{\mathrm{log}}_{a}z\right)\\ -\frac{1}{4}\left({\mathrm{log}}_{a}y-{\mathrm{log}}_{a}z\right)\left({\mathrm{log}}_{a}z-{\mathrm{log}}_{a}y\right)\\ ={\left({\mathrm{log}}_{a}x\right)}^{2}-\left({\mathrm{log}}_{a}y+{\mathrm{log}}_{a}z\right)\left({\mathrm{log}}_{a}x\right)+\frac{1}{4}{\left({\mathrm{log}}_{a}y-{\mathrm{log}}_{a}x\right)}^{2}\\ ={\left({\mathrm{log}}_{a}x-\frac{1}{2}\left({\mathrm{log}}_{a}y+{\mathrm{log}}_{a}z\right)\right)}^{2}\\ \ge 0\end{array}$

よって、問題の 不等式は成り立つ。

等号が成り立つ のは、

$\begin{array}{l}{\mathrm{log}}_{a}x=\frac{1}{2}\left({\mathrm{log}}_{a}y+{\mathrm{log}}_{a}z\right)\\ {\mathrm{log}}_{a}x={\mathrm{log}}_{a}\sqrt{yz}\\ x=\sqrt{yz}\end{array}$

の 場合。

（証明終）

コード

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, log, Rational, sqrt, solve
from sympy.plotting import plot3d

print('38.')

x, y, z = symbols('x, y, z', positive=True)
for a in [Rational(1, 2), 2]:
l = log(x / y, a) * log(x / z, a)
r = log(sqrt(y / z), a) * log(sqrt(z / y, x,), a)
for o in [(l - r).subs({x: 4, y: 2, z: 3}) >= 0, solve(l - r)]:
pprint(o)
print()

p = plot3d(l.subs({z: 2}), r.subs({z: 2}),
(x, 0.1, 10), (y, 0.1, 10), show=False)

p.show()
p.save('sample38.png')

% ./sample38.py
38.
True

[{x: √y⋅√z}]

True

[{x: √y⋅√z}]

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