Bunhiacopxki: \(\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\left(ab+bc+ca\right)\ge\left(a+b+c\right)^2\)
\(\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge\left(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}\right)^2\)
Nhân vế với vế:
\(\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)^2\left(ab+bc+ca\right)\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\left(a+b+c\right)^2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Do: \(\left(ab+bc+ca\right)\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\left(ab+bc+ca\right)\left(\dfrac{a+b+c}{abc}\right)\)
\(=\left(\dfrac{ab+bc+ca}{abc}\right)\left(a+b+c\right)=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\)
\(\text{⇔}\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\ge3+\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
Áp dụng bất đẳng thức AM-GM :
\(\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}\ge3\\ \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge3\)
Ta có :
\(\dfrac{a^2}{b^2}+1\ge\dfrac{2a}{b}\\ \dfrac{b^2}{c^2}+1\ge\dfrac{2b}{c}\\ \dfrac{c^2}{a^2}+1\ge\dfrac{2c}{a}\)
Ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+3\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+3\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+3\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
Suy ra :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\ge3+\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
