Question

数学~~~~~

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Answer 1

b = aq
c = aq^2
q > 0

a^2+b^2+c^2 = a^2 + a^2*q^2 + a^2+q^4
= a^2(1+q^2+q^4)

(a-b+c)^2 = (a-aq+aq^2)^2 = [a(1-q+q^2)]^2 = a^2(1-q+q^2)^2

那么我们只需要比较(1+q^2+q^4)和(1-q+q^2)^2。
所以 (1+q^2+q^4)-(1-q+q^2)^2 = ... = 2q^3-2q^2+2q
所以你只需要证明在q>0的情况下， 2q^3-2q^2+2q > 0 (可以用一次和二次微分证明)

Answer 2

a = +,
b = +,
c = +,
b = aq,
c = bq

a^2+b^2+c^2 & (a-b+c)^2

===================================================
(a-b+c)^2 = ... = a^2 + b^2 + c^2 + 2(ac-bc-ab)
=>
如果ac-bc-ab > 0,
a^2+b^2+c^2 < (a-b+c)^2;
如果ac-bc-ab < 0,
a^2+b^2+c^2 > (a-b+c)^2.

ac-bc-ab = ... = a^2 (q^2 - q^3 - q) = a^2 (q - q^2 - 1)q
要判定q - q^2 - 1 ，
可以用画图/分析 q - q^2 - 1 何时等于0
判别式b-4ac = (1) - 4(-1)(-1) = -3
加上q - q^2 - 1是向下的抛物线，所以q - q^2 - 1在任何时候都小于0

=〉ac-bc-ab = a^2 (q - q^2 - 1)q < 0
=> a^2+b^2+c^2 > (a-b+c)^2.