A
a. có (a-b)2≥0
<=> (a-b)2+4ab≥4ab
<=>(a+b)2≥4ab
<=>\(\sqrt{\left(a+b\right)^2}\ge\sqrt{4ab}\)
<=>a+b≥2\(\sqrt{ab}\)
<=>\(\frac{a+b}{2}\ge\sqrt{ab}\)
b. có (a-b)2≥0
<=> (a-b)2+2ab≥2ab
<=>a2+b2≥2ab
<=>\(\frac{a^2+b^2}{ab}\ge2\)
<=>\(\frac{a}{b}+\frac{b}{a}\ge2\)