Question

giúp e vs ạ

Answer 1

Ta có: \(\frac34\left(a+b\right)^2+\frac14\left(a-b\right)^2\)

\(=\frac34\left(a^2+2ab+b^2\right)+\frac14\left(a^2-2ab+b^2\right)\)

\(=\frac34a^2+\frac32ab+\frac34b^2+\frac14a^2-\frac12ab+\frac14b^2\)

\(=a^2+ab+b^2\) (ĐPCM)

Ta có: \(\frac34\left(a-b\right)^2+\frac14\left(a+b\right)^2\)

\(=\frac34\left(a^2-2ab+b^2\right)+\frac14\left(a^2+2ab+b^2\right)\)

\(=\frac34a^2-\frac32ab+\frac34b^2+\frac14a^2+\frac12ab+\frac14b^2\)

\(=a^2-ab+b^2\)

Ta có: \(a^2+ab+b^2=\frac34\left(a+b\right)^2+\frac14\left(a-b\right)^2\)

=>\(a^2+ab+b^2-\frac34\left(a+b\right)^2=\frac14\left(a-b\right)^2\ge0\forall a,b\)

=>\(a^2+ab+b^2\ge\frac34\left(a+b\right)^2\) (ĐPCM)

Ta có: \(a^2+ab+b^2=\frac34\left(a+b\right)^2+\frac14\left(a-b\right)^2\)

=>\(a^2+ab+b^2-\frac14\left(a-b\right)^2=\frac34\left(a+b\right)^2\ge0\forall a,b\)

=>\(a^2+ab+b^2\ge\frac14\left(a-b\right)^2\) (ĐPCM)

Ta có: \(a^2-ab+b^2=\frac34\left(a-b\right)^2+\frac14\left(a+b\right)^2\)

=>\(a^2-ab+b^2-\frac34\left(a-b\right)^2=\frac14\left(a+b\right)^2\ge0\forall a,b\)

=>\(a^2-ab+b^2\ge\frac34\left(a-b\right)^2\) (ĐPCM)

Ta có: \(a^2-ab+b^2=\frac34\left(a-b\right)^2+\frac14\left(a+b\right)^2\)

=>\(a^2-ab+b^2-\frac14\left(a+b\right)^2=\frac34\left(a-b\right)^2\ge0\forall a,b\)

=>\(a^2-ab+b^2\ge\frac14\left(a+b\right)^2\) (ĐPCM)