Let a,b,c be real numbers such that 1≤a≤b≤c≤4. Find the minimum value of (a−1)2+(ab−1)2+(bc−1)2+(c4−1)2.
Solution — click to reveal
By QM-AM, 4(a−1)2+(ab−1)2+(bc−1)2+(c4−1)2≥4(a−1)+(ab−1)+(bc−1)+(c4−1)=4a+ab+bc+c4−4.By AM-GM, a+ab+bc+c4≥444=42,so 4(a−1)2+(ab−1)2+(bc−1)2+(c4−1)2≥2−1,and (a−1)2+(ab−1)2+(bc−1)2+(c4−1)2≥4(2−1)2=12−82.Equality occurs when a=2,b=2, and c=22, so the minimum value is 12−82.