ta có \(3\le a\le5\)
=> (a-3)(a-5) ≤ 0
<=> a2-5a-3a+15 ≤ 0
<=> a2-8a+15 ≤ 0 (1)
\(3\le b\le5\)
=> (b-3)(b-5) ≤ 0
<=> b2 -8b +15 ≤ 0 (2)
\(3\le c\le5\)
=> (c-3)(c-5) ≤ 0
<=> c2 -8c +15 ≤ 0 (3)
(1)+(2)+ (3)
=> a2+b2+c2 -8a-8b-8c +45 ≤ 0
<=> 50-8(a+b+c)+45 ≤ 0
<=> -8(a+b+c) ≤ -95
<=> a+b+c ≥ \(\dfrac{95}{8}\)
=> Min A= 95/8