急！！！數學---(證明不等式)

1) (ab+a+b+1)(ab+ac+bc+c^2)
= [a(b+1)+(b+1)][a(b+c)+c(b+c)]
= (a+1)(b+1)(a+c)(b+c)

[(a+1)(b+1)(a+c)(b+c)]^2
= (a+1)^2(b+1)^2(a+c)^2(b+c)^2
= (a^2+2a+1)(b^2+2b+1)(a^2+2ac+c^2)(b^2+2bc+c^2)
= [(a^2-2a+1)+4a][(b^2-2b+1)+4b][(a^2-2ac+c^2)+4ac][(b^2-2bc+c^2)+4bc]
= [(a-1)^2+4a][(b-1)^2+4b][(a-c)^2+4ac][(b-c)^2+4bc]

As x^2 >= 0 for any x, (a-1)^2 and (b-1)^2 >=0
As a,b,c are not all equal, (a-c)^2 or (b-c)^2 > 0
.'. [(a-1)^2+4a][(b-1)^2+4b][(a-c)^2+4ac][(b-c)^2+4bc] > (4a)(4b)(4ac)(4bc)
(4a)(4b)(4ac)(4bc) = (16abc)^2

[(ab+a+b+1)(ab+ac+bc+c^2)]^2 > (16abc)^2

As a,b,c are all > 0
(ab+a+b+1)(ab+ac+bc+c^2) > 0 and 16abc > 0

(ab+a+b+1)(ab+ac+bc+c^2) > 16abc

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2) (a+b)(a-b)^2 + (a+c)(a-c)^2 + (b+c)(b-c)^2
= (a^2-b^2)(a-b) + (a^2-c^2)(a-c) + (b^2-c^2)(b-c)
= (a^3 - a^2b - ab^2 + b^3) +(a^3 - a^2c - ac^2 + c^3) + (b^3 - b^2c - bc^2 + c^3)
= 2a^3 + 2b^3 + 2c^3 - (a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2)
= 2(a^3 + b^3 + c^3) - [a^2(b+c) + b^2(a+c) + c^2(a+b)]

As a,b,c are not all equal, (a-b)^2, (a-c)^2 or (b-c)^2 > 0
As a,b,c are all > 0, (a+b), (a+c) and (b+c) > 0
.'. (a+b)(a-b)^2 + (a+c)(a-c)^2 + (b+c)(b-c)^2 > 0
.'. 2(a^3 + b^3 + c^3) - [a^2(b+c) + b^2(a+c) + c^2(a+b)] > 0
2(a^3 + b^3 + c^3) > [a^2(b+c) + b^2(a+c) + c^2(a+b)]

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3) a^2 + b^2
= (a^2 - 2ab + b^2) + 2ab
= (a - b)^2 + 2ab
As a, b, c are lengths of 3 sides of a triangle, a,b,c all > 0 and a^2 + b^2 >= c^2
.'. (a - b)^2 + 2ab >= c^2

As a, b > 0, (a-b)^2 >= 0
.'. 2ab >= c^2

Similarly, 2bc >= a^2 and 2ca >= b^2
As a, b, c are lengths of 3 sides of a triangle,
at most oneof [2ab >= c^2, 2bc >= a^2, 2ca >= b^2] is equal

.'. 2ab + 2bc + 2ca > c^2 + a^2 + b^2
.'. 2(ab+bc+ca) > a^2 + b^2 + c^2

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4) 假設矩形的長=a,寬=b
.'. 面積 = ab

假設 L = a+2b
.'. b = (L-a) / 2

面積 = a(L-a) / 2
面積最大當 a = L-a
.'. L = 2a
b = (L-a) / 2
= a / 2
= L/4

矩形的長=L/2,寬=L/4