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Bài 1:
$(a+b+c)^2=(a+b)^2+2(a+b)c+c^2=a^2+2ab+b^2+2ac+2bc+c^2=3a^2+3b^2+3c^2$
$\leftrightarrow 2ab+2ac+2bc=2a^2+2b^2+2c^2$
$\leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2=0$
$\leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0$
$\leftrightarrow$ $a-b=0 \leftrightarrow a=b(1) \\ a-c=0 \leftrightarrow a=c(2)$
Từ (1) và (2) suy ra dpcm
Bài 2:
$a+b+c=0 \leftrightarrow (a+b+c)^2=0$
$ \leftrightarrow a^2+b^2+c^2+2(ab+ac+bc)=0$
$\leftrightarrow a^2+b^2+c^2=-2(ab+ac+bc)$
$\leftrightarrow (a^2+b^2+c^2)^2=4(ab+ac+bc)^2=4a^2b^2+4a^2c^2+4b^2c^2+8(a^2bc+b^2ac+c^2ab)$
$\leftrightarrow a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)=4a^2b^2+4a^2c^2+4b^2c^2+8abc(a+b+c$
$\leftrightarrow a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2$
Mặt khác ta có:
$2(ab+bc+ca)^2=2a^2b^2+2b^2c^2+2a^2c^2+4abc(a+b+c)=$$a^2b^2+b^2c^2+a^2c^2$

Vì vậy $a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2=2(ab+bc+ca)^2$
Vậy $a^4+b^4+c^4=2(ab+ac+bc)^2$

 
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