Có: `(a+b+c)^2=a^2+b^2+c^2.`
`⇔a^2+b^2+c^2+2(ac+bc+ca)=a^2+b^2+c^2`
`⇔a^2+b^2+c^2+2(ac+bc+ca)-a^2-b^2-c^2==0`
`⇔2(ab+bc+ca)=0`
`⇔ab+bc+ca=0`
`⇒ab=-(bc+ca); bc=-(ca+ab); ca=-(ab+bc)`
Có: `P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}.`
`⇔P=\frac{a^2}{a^2+bc+bc}+\frac{b^2}{b^2+ac+ac}+\frac{c^2}{c^2+ab+ab}.`
Thay `ab=-(bc+ca); bc=-(ca+ab); ca=-(ab+bc)` vào `P` ta có:
`⇔P=\frac{a^2}{a^2+bc-(ca+ab)}+\frac{b^2}{b^2+ac-(ab+bc)}+\frac{c^2}{c^2+ab-(bc+ca)}`
`⇔P=\frac{a^2}{a^2+bc-ca-ab}+\frac{b^2}{b^2+ac-ab-bc}+\frac{c^2}{c^2+ab-bc-ca}`
`⇔P=\frac{a^2}{(a^2-ca)+(bc-ab)}+\frac{b^2}{(-ab+b^2)+(ac-bc)}+\frac{c^2}{(-bc+ab)+(c^2-ca)}`
`⇔P=\frac{a^2}{a(a-c)-b(a-c)}+\frac{b^2}{-b(a-b)+c(a-b)}+\frac{c^2}{b(a-c)-c(a-c)}`
`⇔P=\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(a-b)(c-b)}+\frac{c^2}{(a-c)(b-c)}`
`⇔P=\frac{a^2}{(a-b)(a-c)}-\frac{b^2}{(a-b)(b-c)}+\frac{c^2}{(a-c)(b-c)}`
`⇔P=\frac{a^2(b-c)-b^2(a-c)+c^2(a-b)}{(a-b)(a-c)(b-c)}`
`⇔P=\frac{a^2(b-c)-b^2a+b^2c+c^2a-c^2b}{(a-b)(a-c)(b-c)}`
`⇔P=\frac{a^2(b-c)+(c^2a-b^2a)+(b^2c-c^2b)}{(a-b)(a-c)(b-c)}`
`⇔P=\frac{a^2(b-c)-a(b-c)(b+c)+bc(b-c)}{(a-b)(a-c)(b-c)}`
`⇔P=\frac{(b-c)(a^2-ab-ac+bc)}{(a-b)(a-c)(b-c)}`
`⇔P=\frac{(b-c)[a(a-b)-c(a-b)]}{(a-b)(a-c)(b-c)}`
`⇔P=\frac{(b-c)(a-c)(a-b)}{(a-b)(a-c)(b-c)}`
`⇔P=1.`
Vậy `P=1.`
