Question

chứng minh rằng

\(\left(a^2+b^2\right)^2\ge8\left(a-b\right)^2\)

với a>b,ab=1

Answer 1

\(\left(a^2+b^2\right)^2=\left[\left(a-b\right)^2+2ab\right]^2=\left[\left(a-b\right)^2+2\right]^2\ge8\left(a-b\right)^2\) (đpcm)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a-b=\sqrt{2}\\ab=1\end{matrix}\right.\) \(\Rightarrow\left(a;b\right)=\left(\dfrac{\sqrt{6}+\sqrt{2}}{2};\dfrac{\sqrt{6}-\sqrt{2}}{2}\right);\left(\dfrac{\sqrt{2}-\sqrt{6}}{2};-\dfrac{\sqrt{2}+\sqrt{6}}{2}\right)\)