Question

(a^2-b^2)(c^2-d^2)>_ (ac-bd)^2

Answer 1

Thử sức xíu, có j sai mong bỏ qua, xie xie :3

\(VT=\left(ac\right)^2-\left(ad\right)^2-\left(bc\right)^2+\left(bd\right)^2\)

\(=\left(ac\right)^2+\left(bd\right)^2-\left[\left(ad\right)^2+\left(bc\right)^2\right]\)

Có \(\left(ad-bc\right)^2\ge0\Rightarrow\left(ad\right)^2+\left(bc\right)^2\ge2abcd\)

\(\Rightarrow\left(ac\right)^2+\left(bd\right)^2-\left[\left(ad\right)^2+\left(bc\right)^2\right]\ge\left(ac\right)^2+\left(bd\right)^2-2abcd=\left(ac-bd\right)^2\)