a, Ta xét hiệu \(\frac{a^2+b^2}{2}-\left(\frac{a+b}{2}\right)^2\)
\(=\frac{2\left(a^2+b^2\right)}{4}-\frac{a^2+2ab+b^2}{4}=\frac{1}{4}\left(2a^2+2b^2-a^2-b^2-2ab\right)=\frac{1}{4}\left(a+b\right)^2\ge0\)
Vậy \(\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2.\)
Dấu "="xảy ra khi a = b.
b, Ta xét hiệu:
\(\frac{a^2+b^2+c^2}{3}-\left(\frac{a+b+c}{3}\right)^2=\frac{1}{9}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)
Vậy \(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\)
Dấu "=" xảy ra khi a = b = c.
Cách khác:\(a^2+b^2\ge2ab\Leftrightarrow a^2+b^2+a^2+b^2\ge\left(a+b\right)^2\Leftrightarrow a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)
\(\frac{a^2+b^2}{2}\ge\frac{\frac{\left(a+b\right)^2}{2}}{2}=\left(\frac{a+b}{2}\right)^2\)
Dấu " = " xảy ra <=> a=b
b) \(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)( tự c/m)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow\frac{\left(a^2+b^2+c^2\right)}{3}\ge\frac{\left(a+b+c\right)^2}{9}=\left(\frac{a+b+c}{3}\right)^2\)
Dấu " = " xảy ra <=> a=b=c
