Question

Cho a, b, c thỏa mãn: \(\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=1\). Tính \(\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}\) và CMR: \(ab+bc+ca\le3\)

Answer 1

\(\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}=\dfrac{a^2+2-2}{a^2+2}+\dfrac{b^2+2-2}{b^2+2}+\dfrac{c^2+2-2}{c^2+2}\)

\(=3-2\left(\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}\right)=3-2=1\)

Từ đó:

\(1=\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+6}\)

\(\Rightarrow a^2+b^2+c^2+6\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow ab+bc+ca\le3\) (đpcm)