BĐT Cosi cho 2 số a,b >0:
a + b >= 2căn(ab)
di từ: ( √a - √b)² ≥ 0 ( voi moi a , b ≥ 0 )
<=> a + b - 2√(ab) ≥ 0
<=> a + b ≥ 2√(ab)
dau "=" xay ra khi √a - √b = 0 <=> a = b
Ta có:\(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)\(\forall a,b\ge0\)
\(\Leftrightarrow a+b-2\sqrt{ab}\ge0\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\left(đpcm\right)\)
\(a+b\ge2\sqrt{ab}\)
\(\Leftrightarrow a+b-2\sqrt{ab}\ge0\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge o\)(Luôn đúng)