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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a^2+1=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)
Tương tự: \(\left\{{}\begin{matrix}b^2+1=\left(a+b\right)\left(b+c\right)\\c^2+1=\left(c+a\right)\left(b+c\right)\end{matrix}\right.\)
=> \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
Mặt khác: \(a+b+c-abc=a\left(1-bc\right)+b+c\)
\(=a\left(ab+ca\right)+b+c\) (Vì ab+bc+ca=1)
\(=\left(a^2+1\right)\left(b+c\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (Vì \(a^2+1=\left(a+b\right)\left(c+a\right)\))
\(T=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sum\dfrac{a}{b^2+bc+c^2}\ge\dfrac{\left(a+b+c\right)^2}{ab^2+abc+ac^2+bc^2+abc+ba^2+ca^2+abc+cb^2}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}=\dfrac{a+b+c}{ab+bc+ac}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3x^2y+y^3\)
\(=6x^2y+2y^3\)
\(=2y\left(3x^2+y^2\right)\)
Lời giải:
$a+bc=a(a+b+c)+bc=(a+b)(a+c)$
Tương tự: $b+ca=(b+a)(b+c); c+ab=(c+a)(c+b)$
Do đó:
$P=\frac{b-c}{(a+b)(a+c)}+\frac{c-a}{(b+a)(b+c)}+\frac{a-b}{(c+a)(c+b)}$
$=\frac{(b-c)(b+c)+(c-a)(c+a)+(a-b)(a+b)}{(a+b)(b+c)(c+a)}$
$=\frac{b^2-c^2+c^2-a^2+a^2-b^2}{(a+b)(b+c)(c+a)}$
$=\frac{0}{(a+b)(b+c)(c+a)}=0$