Question

Answer 1

\(a^2+b^2+c^2+a=a^2+b^2+c^2+a\left(a+b+c\right)\)

\(=2a^2+b^2+c^2+ab+ac=2a^2+b\left(a+b\right)+c\left(a+c\right)\)

\(\Rightarrow\dfrac{1}{a^2+b^2+c^2+a}=\dfrac{\left(a+b+c\right)^2}{2a^2+b\left(a+b\right)+c\left(a+c\right)}\le\dfrac{a^2}{2a^2}+\dfrac{b^2}{b\left(a+b\right)}+\dfrac{c^2}{c\left(a+c\right)}\)

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

Tương tự và cộng lại:

\(VT\le\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{3}{2}=3+\dfrac{3}{2}=\dfrac{9}{2}\) (đpcm)