Question

(abc^(2))/(a+1) .(c)/(a^(2)b^(2))

Answer 1

Answer:

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Answer 2

Answer:

Notice that for all real a,b,c,

(a−1)2+(b−1)2+(c−1)2≥0

a2+b2+c2−2a−2b−2c+3≥0

a2+b2+c2≥−3+(a+b+c)+(a+b+c)

But by AM-GM, a+b+c≥3abc−−−√3=3. So,

a2+b2+c2≥−3+3+(a+b+c)

a2+b2+c2≥a+b+c

Equality is attained when a=b=c=1.