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Ta có
(m+n+p)^q >= m^q+n^q+p^q
=>a+b+c=1
=>(a+b+c)^2016=1 >= a2016 + b2016 + c2016
Mà a2016 + b2016 + c2016 >=0
=> a2016 + b2016 + c2016=1
\(201^2=\left(200+1\right)^2=200^2+2.200.1+1^2=40000+400+1=40401\)
\(498^2=\left(500-2\right)^2=500^2-2.500.2+2^2=250000-2000+4=248004\)
a2+b2+c2=ab+bc+ca
<=>2a2+2b2+2c2=2ab+2bc+2ca
<=>(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ca+a2)=0
<=>(a-b)2+(b-c)2+(c-a)2=0
<=>a=b=c
mà a+b+c=3<=>a=b=c=1
=>P=0