Step-by-step explanation:

We can use the area formula for a triangle in terms of its side lengths, known as Heron's formula, to solve this problem. Let s = (a + b + c)/2 be the semiperimeter of the triangle, then the area of the triangle is given by:

A = √(s(s-a)(s-b)(s-c))

Since A = 1, we have:

1 = √(s(s-a)(s-b)(s-c))

Squaring both sides, we get:

s(s-a)(s-b)(s-c) = 1

Expanding the product, we get:

s^4 - (a+b+c)s^3 + (ab+bc+ca)s^2 - abc(s) = 1

Since s = (a+b+c)/2, we can substitute to get:

[(a+b+c)/2]^4 - (a+b+c)^3/2 + (ab+bc+ca)(a+b+c)^2/4 - abc(a+b+c)/2 - 1 = 0

Simplifying and factoring out (a+b+c), we get:

(a+b+c)[(a+b+c)^3/16 - (a+b+c)/2 + (ab+bc+ca)/4 - abc/2 - 1/(a+b+c)] = 0

Since a, b, c are positive, we have a+b+c > 0. Therefore, the second factor must be equal to zero:

(a+b+c)^3/16 - (a+b+c)/2 + (ab+bc+ca)/4 - abc/2 - 1/(a+b+c) = 0

Multiplying both sides by 16(a+b+c), we get:

(a+b+c)^4 - 8(a+b+c)^2 + 16(ab+bc+ca)(a+b+c) - 32abc(a+b+c) - 16 = 0

We can use the fact that a ≥ b ≥ c to simplify this expression. Since (a-b)(a-c) ≥ 0, we have:

ab + ac ≥ b^2 + bc

2(ab + ac) ≥ b^2 + 2bc

Substituting this inequality into the previous expression, we get:

(a+b+c)^4 - 8(a+b+c)^2 + 2(b^2+2bc+c^2)(a+b+c) - 32abc(a+b+c) - 16 ≤ 0

Simplifying and factoring, we get:

(b-c)^2(a+b+c-2√2)(a+b+c+2√2) ≤ 0

Since b ≥ c, we have b-c ≥ 0. Therefore, the only way for this inequality to hold is if a+b+c ≥ 2√2. Since the area of the triangle is fixed at 1, we have:

1 = √(s(s-a)(s-b)(s-c)) ≤ √(s(s-c)^3) = √[(a+b+c)(a+b-c)(a+c-b)(b+c-a)] ≤ √[(a+b+c)(a+b+c-2b)(a+b+c-2c)(a+b+c-2a)]

= √[(a+b+c)(a+b+c-4b)(a+b+c-4c)(a+b+c-4a)]

= √[(a+b+c)(3a-3b+c)(3a+3b-5c)(-3a+b+3c)]

Since a, b, c are positive, we have a+b+c > 0. Therefore, the only way for this inequality to hold is if b ≥

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