\(a^2+b^2+c^2\ge a(b+c)\)
\(\to 2a^2+2b^2+2c^2\ge 2ab+2ac\)
\(\to 2a^2+2b^2+2c^2-2ab-2ac\ge 0\)
\(\to (a^2-2ab+b^2)+(a^2-2ac+c^2)+b^2+c^2\ge 0\)
\(\to (a-b)^2+(a-c)^2+b^2+c^2\ge 0\) (luôn đúng)
\(\to\) Dấu "=" xảy ra khi \(\begin{cases}a-b=0\\a-c=0\\b=0\\c=0\end{cases}\to a=b=c=0\)