Question

cho a,b,c > 0 thỏa mãn a+b+c=3 Chứng minh rằng \(\sum\)\(\dfrac{a}{a^2+b+c}\) ≤ 1

Answer 1

Bunhiacopxki: \(\left(a^2+b+c\right)\left(1+b+c\right)\ge\left(a+b+c\right)^2=9\)

\(\Rightarrow\dfrac{1}{a^2+b+c}\le\dfrac{1+b+c}{9}\)

\(\Rightarrow\dfrac{a}{a^2+b+c}\le\dfrac{a+ab+ac}{9}\)

Tương tự: \(\dfrac{b}{b^2+a+c}\le\dfrac{b+ab+bc}{9}\) ; \(\dfrac{c}{c^2+a+b}\le\dfrac{c+ac+bc}{9}\)

Cộng vế:

\(P\le\dfrac{a+b+c+2\left(ab+bc+ca\right)}{9}\le\dfrac{a+b+c+\dfrac{2}{3}\left(a+b+c\right)^2}{9}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)