Let a, b and c be real numbers such that a+2b+c = 4. Find the maximum value of (ab +bc + ca).
Given,
a+2b+c=4⇒a=4−2b−clet y=ab+bc+ca=a(b+c)+bc =(4−2b−c)(b+c)+bc =−2b2+4b−2bc+4c−c22b2+2(c−2)b−4c+c2+y=0
Since b∈R,so
4(c−2)2−4×2×(−4c+c2+y)≥0
⇒(c−2)2+8c−2c2−2y≥0⇒c2−4c+2y−4≤0
Since c∈R so,16−4(2y−4)≥0⇒y≤4
Hence, maximum value of ab + bc + ca is 4.