Ta có \(\left(b+c-a\right)^2+2ab+2ac-2bc=a^2+b^2+c^2\ge0\)
\(\Leftrightarrow\left(b+c-a\right)^2\ge2bc-2ab-2ac\)
Áp dụng BĐT Bunhiacopski: \(2\left(b^2+c^2\right)\ge\left(b+c\right)^2\)
\(\dfrac{\left(b+c-a\right)^2}{2a^2+\left(b+c\right)^2}\ge\dfrac{2bc-2ac-2ab}{2a^2+2\left(b^2+c^2\right)}=\dfrac{bc-ac-ab}{a^2+b^2+c^2}\)
Cmtt \(\dfrac{\left(c+a-b\right)^2}{2b^2+\left(c+a\right)^2}\ge\dfrac{ac-ab-bc}{a^2+b^2+c^2};\dfrac{\left(a+b-c\right)^2}{2c^2+\left(a+b\right)^2}\ge\dfrac{ab-ac-bc}{a^2+b^2+c^2}\)
Cộng VTV
\(\Leftrightarrow VT\ge\dfrac{bc-ac-ab+ac-ab-bc+ab-ac-bc}{a^2+b^2+c^2}=\dfrac{-\left(ab+bc+ca\right)}{a^2+b^2+c^2}\)
Ta thấy \(\left(a+b+c\right)^2\ge0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow-\left(ab+bc+ca\right)\le\dfrac{a^2+b^2+c^2}{2}\Leftrightarrow ab+bc+ca\ge-\dfrac{a^2+b^2+c^2}{2}\)
Do đó \(VT\ge\dfrac{\dfrac{a^2+b^2+c^2}{2}}{a^2+b^2+c^2}=\dfrac{1}{2}\)
Dấu \("="\Leftrightarrow a=b=c\)
