b1: ta có: a^2+b^2 >0 ; b^2 +c^2>0 ; c^2 +a^2>0
=> \(a^2+b^2\ge2\sqrt{a^2.b^2}\) (BĐT cau chy)
\(b^2+c^2\ge2\sqrt{b^2.c^2}\) (BĐT cau chy)
\(c^2+a^2\ge2\sqrt{c^2.a^2}\)(BĐT cauchy)
=>\(\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\ge8a^2.b^2.c^2\)
Dấu '= xảy ra khi a=b=c (đpcm)
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