\(\dfrac{1}{a^2+b^2+2}+\dfrac{1}{b^2+c^2+2}+\dfrac{1}{c^2+a^2+2}\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a^2+b^2}{a^2+b^2+2}+\dfrac{b^2+c^2}{b^2+c^2+2}+\dfrac{c^2+a^2}{c^2+a^2+2}\ge\dfrac{3}{2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT\ge\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6}\)
\(\ge\dfrac{\sqrt{3\left(a^2b^2+b^2c^2+a^2c^2\right)}+2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}\)
\(\ge\dfrac{2\left(a^2+b^2+c^2\right)+ab+bc+ca}{a^2+b^2+c^2}\)
Cần chứng minh \(\dfrac{2\left(a^2+b^2+c^2\right)+ab+bc+ca}{a^2+b^2+c^2}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge0\) *luôn đúng*
Em có cách khác :v
\(\dfrac{1}{a^2+b^2+2}\le\dfrac{1}{\dfrac{\left(a+b\right)^2}{2}+2}=\dfrac{1}{\dfrac{\left(3-c\right)^2}{2}+2}\\ =\dfrac{2}{\left(3-c\right)^2+4}=\dfrac{2}{c^2-6c+13}\)
Ta cần CM:
\(\dfrac{2}{c^2-6c+13}\le\dfrac{1}{8}c+\dfrac{1}{8}\\ \Leftrightarrow\left(3-c\right)\left(c-1\right)^2\ge0\left(luon;dung\right)\\ \Rightarrow A\le\dfrac{1}{8}a+\dfrac{1}{8}+\dfrac{1}{8}b+\dfrac{1}{8}+\dfrac{1}{8}c+\dfrac{1}{8}=\dfrac{3}{4}\)
Nguồn : Anh hùng
