Question

cho \(a^2+b^2=6\) ( a, b dương)

chứng minh : \(\sqrt{3\left(a^2+6\right)}\ge\sqrt{2}\left(a+b\right)\)

Answer 1

\(\sqrt{3\left(a^2+6\right)}\ge\sqrt{2}\left(a+b\right)\)

\(\Leftrightarrow\sqrt{3\left(2a^2+b^2\right)}\ge\sqrt{2}\left(a+b\right)\)

\(\Leftrightarrow6a^2+3b^2\ge2a^2+4ab+2b^2\)

\(\Leftrightarrow4a^2-4ab+b^2\ge0\)

\(\Leftrightarrow\left(2a-b\right)^2\ge0\)(đúng)

=> ĐPCM