@@
Đặt: \(A=\frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}\)
\(2A=\frac{2a}{2a+b}+\frac{2b}{2b+c}+\frac{2c}{2c+a}\)
\(2A-3=\frac{2a}{2a+b}-1+\frac{2b}{2b+c}-1+\frac{2c}{2c+a}-1\)
\(2A-3=\frac{-b}{2a+b}+\frac{-c}{2b+c}+\frac{-a}{2c+a}\)
cần cm: \(A\le1\) hay \(2A-3\le-1\) hay
\(\frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}\ge1\)
bđt này hiển nhiên đúng theo cauchy-schwarz:
\(\frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}=\frac{b^2}{2ab+b^2}+\frac{c^2}{2bc+c^2}+\frac{a^2}{2ac+a^2}\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)
Vậy bđt được chứng minh @@. "=" khi a=b=c
\(gt\Leftrightarrow\sum\frac{2a}{2a+b}\le2\Leftrightarrow\sum\frac{b}{2a+b}\ge1\)
ta cần cm bđt trên đúng. Thật vậy:
\(\sum\frac{b}{2a+b}=\sum\frac{b^2}{2ab+b^2}\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)(đúng)
\("="\Leftrightarrow a=b=c\)
