Question

Cho 3 số dương \(0\le a\le b\le c\le1\) chứng minh rằng: \(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le2\)

Answer 1

\(a\le1;b\le1\Rightarrow a-1\le0;b-1\le0\)

\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)

\(\Leftrightarrow ab-a-b+1\ge0\)

\(\Leftrightarrow ab+1\ge a+b\)

\(\frac{1}{ab+1}\le\frac{1}{a+b}\)

\(\Rightarrow\frac{c}{ab+1}\le\frac{2c}{a+b+c}\)

Chứng minh tương tự ta cũng có :

\(\hept{\begin{cases}\frac{a}{bc+1}\le\frac{2a}{a+b+c}\\\frac{b}{ac+1}\le\frac{2b}{a+b+c}\end{cases}}\)

Cộng vế với vế ta được :

\(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2a+2b+2c}{a+b+c}\)

\(\Leftrightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2\left(a+b+c\right)}{a+b+c}=2\)  (đpcm)