第 2 題
若\(ab+bc+ca=3\),\(a,b,c>0\),試證明\(\displaystyle \frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}\le 1\)。
之前寫過
原式即證明
\(\displaystyle \frac{1}{2}-\frac{1}{a^2+2}+\frac{1}{2}-\frac{1}{b^2+2}+\frac{1}{2}-\frac{1}{c^2+2}\ge \frac{1}{2}\)
\(\displaystyle \frac{a^2}{2(a^2+2)}+\frac{b^2}{2(b^2+2)}+\frac{c^2}{2(c^2+2)}\ge \frac{1}{2}\)
\(\displaystyle \frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\ge 1\)
\(\displaystyle \left(\frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\right)(a^2+2+b^2+2+c^2+2)\ge(a+b+c)^2\)
\(\displaystyle \frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2+6}=1\)