Question

\(\sqrt{a}+\sqrt{b}\)  ≥  \(\sqrt{a+b}\)       (a , b ≥ 0)

chúng minh ạ !!!

 

Answer 1

Ta có:\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)

    \(\Leftrightarrow a+2\sqrt{ab}+b\ge a+b\)

    \(\Leftrightarrow2\sqrt{ab}\ge0\Leftrightarrow\sqrt{ab}\ge0\) (luôn đúng do a,b≥0)

Answer 2

\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)

\(< =>a+2\sqrt{ab}+b\ge a+b\)

\(< =>2\sqrt{ab}\ge0\) (luôn đúng \(\forall a,b\ge0\))

dấu"=" xảy ra<=>a=b=0

Answer 3

Ta có: \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)

\(\Leftrightarrow a+2\sqrt{ab}+b\ge a+b\)

\(\Leftrightarrow2\sqrt{ab}\ge0\)(luôn đúng