\(\Leftrightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}-7\le0\)
Đặt \(P=\frac{a}{c}+\frac{c}{a}+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}-7\)
Không mất tỉnh tổng quát, giả sử \(a\le b\le c\Rightarrow\left(a-b\right)\left(b-c\right)\ge0\)
\(\Rightarrow ab+bc\ge b^2+ac\Rightarrow\left\{{}\begin{matrix}\frac{a}{c}+1\ge\frac{b}{c}+\frac{a}{b}\\1+\frac{c}{a}\ge\frac{b}{a}+\frac{c}{b}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{c}+\frac{c}{a}+2\ge\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}\)
\(\Rightarrow P\le\frac{a}{c}+\frac{c}{a}+\frac{a}{c}+\frac{c}{a}+2-7=2\left(\frac{a}{c}+\frac{c}{a}\right)-5\)
Do \(1\le a\le c\le2\Rightarrow1\le\frac{c}{a}\le2\)
Đặt \(\frac{c}{a}=x\Rightarrow1\le x\le2\)
\(\Rightarrow P\le2\left(x+\frac{1}{x}\right)-5=\frac{2x^2-5x+2}{x}=\frac{\left(2x-1\right)\left(x-2\right)}{x}\le0\) (đpcm)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;1;2\right);\left(1;2;2\right)\) và các hoán vị
=\(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)
=3+\(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
áp dụng hệ quả của bđt côsi \(\frac{a}{b}+\frac{b}{a}\ge2\)với a,b >0 ta có BĐT cuối cùng luôn đúng
vậy .....
