Question

Answer 1

\(VT=\dfrac{a}{3-bc}+\dfrac{b}{3-ca}+\dfrac{c}{3-ab}\le\dfrac{a}{3-\dfrac{b^2+c^2}{2}}+\dfrac{b}{3-\dfrac{c^2+a^2}{2}}+\dfrac{c}{3-\dfrac{a^2+b^2}{2}}\)

\(VT\le\dfrac{2a}{a^2+3}+\dfrac{2b}{b^2+3}+\dfrac{2c}{c^2+3}\)

Ta có:

\(a^2+1+1+1\ge4\sqrt[4]{a^2}=4\sqrt{a}\)

\(\Rightarrow\left(a^2+3\right)^2\ge16a\)

\(\Rightarrow\dfrac{2a}{a^2+3}\le\dfrac{a^2+3}{8}\)

Tương tự và cộng lại:

\(VT\le\dfrac{a^2+3}{8}+\dfrac{b^2+3}{8}+\dfrac{c^2+3}{8}=\dfrac{3}{2}\) (đpcm)