a. Ta có: \(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow a^2+b^2+1-ab-a-b\ge0\)
\(\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b\ge0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2b+1+a^2-2a+1\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-1\right)^2+\left(a-1\right)^2\ge0\left(1\right)\)
Ta có: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a;b\\\left(b-1\right)^2\ge0\forall b\\\left(a-1\right)^2\ge0\forall a\end{matrix}\right.\)
=> \(\left(a-b\right)^2+\left(b-1\right)^2+\left(a-1\right)^2\ge0\)
=> Đẳng thức (1) luôn đúng
=> đpcm
b. Ta có: \(a^2+b^2+c^2\ge a\left(b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ac-ab\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc\ge0\)
\(\Leftrightarrow a^2-2ab+b^2+a^2-2ac+c^2+b^2+c^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+c^2+b^2\ge0\)(luôn đúng)
=> đpcm.
