thay đề liên tục nhỉ
\(\sum\dfrac{1}{a}=\dfrac{\left(b^2+c^2\right)}{a\left(b^2+c^2\right)}+\dfrac{a^2+c^2}{b\left(a^2+c^2\right)}+\dfrac{b^2+a^2}{c\left(b^2+a^2\right)}\)
\(=\dfrac{b^2}{a\left(b^2+c^2\right)}+\dfrac{c^2}{a\left(b^2+c^2\right)}+\dfrac{a^2}{b\left(a^2+c^2\right)}+\dfrac{c^2}{b\left(a^2+c^2\right)}+\dfrac{b^2}{c\left(a^2+b^2\right)}+\dfrac{a^2}{c\left(a^2+b^2\right)}\)
=\(\sum\left(\dfrac{a^2}{b\left(a^2+c^2\right)}+\dfrac{b^2}{a\left(b^2+c^2\right)}\right)\ge\sum\dfrac{\left(a+b\right)^2}{b\left(a^2+c^2\right)+a\left(b^2+c^2\right)}\) cauchy shawrtz
\(=\sum\dfrac{\left(a+b\right)^2}{a^2b+bc^2+ab^2+ac^2}=\sum\dfrac{\left(a+b\right)^2}{\left(a+b\right)\left(ab+c^2\right)}\)
\(=\sum\dfrac{a+b}{ab+c^2}\)(Q.E.D)
@Vũ Tiền Châu @Akai Haruma @Mysterious Person @Phùng Khánh Linh
Lời giải:
Áp dụng Cauchy-Schwarz kết hợp AM-GM
\(\frac{a+b}{bc+a^2}=\frac{(a+b)(b+c)}{(bc+a^2)(b+c)}=\frac{(a+b)(b+c)}{b(a^2+c^2)+c(a^2+b^2)}\)
\(\leq \frac{1}{2}\frac{(a+b)^2+(b+c)^2}{b(a^2+c^2)+c(a^2+b^2)}=\frac{1}{2}\left(\frac{(a+b)^2}{b(a^2+c^2)+c(a^2+b^2)}+\frac{(b+c)^2}{b(a^2+c^2)+c(a^2+b^2)}\right)\)
\(\leq \frac{1}{2}\left(\frac{a^2}{b(a^2+c^2)}+\frac{b^2}{c(a^2+b^2)}+\frac{b^2}{c(a^2+b^2)}+\frac{c^2}{b(a^2+c^2)}\right)\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(\sum \frac{a+b}{bc+a^2}\leq \frac{a^2+c^2}{b(a^2+c^2)}+\frac{b^2+a^2}{c(a^2+b^2)}+\frac{c^2+b^2}{a(b^2+c^2)}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
