Question

chứng minh

\(\sqrt{\frac{a+b}{2}}\) >= \(\frac{\sqrt{a}+\sqrt{b}}{2}\)

Answer 1

Lời giải:

Áp dụng BĐT Cô-si cho số không âm:

\(a+b\geq 2\sqrt{ab}\)

\(\Rightarrow 2(a+b)\geq a+b+2\sqrt{ab}=(\sqrt{a}+\sqrt{b})^2\)

\(\Rightarrow \frac{a+b}{2}\geq \frac{(\sqrt{a}+\sqrt{b})^2}{4}\)

\(\Rightarrow \sqrt{\frac{a+b}{2}}\geq \sqrt{\frac{(\sqrt{a}+\sqrt{b})^2}{4}}=\frac{\sqrt{a}+\sqrt{b}}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a+b=2\sqrt{ab}\Leftrightarrow (\sqrt{a}-\sqrt{b})^2=0\Leftrightarrow a=b\)