Giải thích các bước giải:
Đề \(\leftrightarrow (a+b)^{2}+(b+c)^{2}+(a+b)(b+c) \leq \frac{a(a+b)(b+c)}{b}+\frac{b(a+b)(b+c)}{c}+\frac{c(a+b)(b+c)}{a}\)
\(\leftrightarrow a^{2}+ac+c^{2}+3b^{2}+3ab+3bc \leq \frac{a^{2}c}{b}+a^{2}+ab+ac+\frac{b^{2}(a+b)}{c}+b^{2}+ab+c^{2}+bc+\frac{bc(b+c)}{a}\)
\(\leftrightarrow 2b^{2}+2bc+ab \leq \frac{a^{2}c}{b}+\frac{b^{2}(b+a)}{c}+\frac{bc(b+c)}{a}\)
VT=\(\frac{a^{2}c}{b}+\frac{b^{2}(b+a)}{c}+\frac{bc(b+c)}{a}=\frac{1}{2}(\frac{a^{2}c}{b}+\frac{b^{3}}{c})+\frac{1}{2}(\frac{a^{2}c}{b}+\frac{bc^{2}}{a})+\frac{1}{2}(\frac{b^{3}}{c}+\frac{bc^{2}}{a})+b^{2}(\frac{c}{a}+\frac{a}{c}) \geq ab+\sqrt{ac^{3}}+\sqrt{\frac{a^{4}c}{a}}+2b^{2} \geq ab+2bc+2b^{2}\)=VP
Dấu "=" xảy ra khi \(a=b=c\)
