Question

Tìm GTNN của A=x+\(\sqrt{x^{2^{ }}+\dfrac{1}{x}}\)

với x>0

Answer 1

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((x^2+\frac{1}{x})(1+8)\geq (x+\frac{2\sqrt{2}}{\sqrt{x}})^2\)

\(\Rightarrow \sqrt{x^2+\frac{1}{x}}\geq \frac{x+\frac{2\sqrt{2}}{\sqrt{x}}}{3}\)

Do đó: \(A\geq x+\frac{x+\frac{2\sqrt{2}}{\sqrt{x}}}{3}=\frac{4x}{3}+\frac{2}{3}\sqrt{\frac{2}{x}}\)

Áp dụng BĐT AM-GM:

\(\frac{4x}{3}+\frac{2}{3}\sqrt{\frac{2}{x}}=\frac{4x}{3}+\frac{1}{3}\sqrt{\frac{2}{x}}+\frac{1}{3}\sqrt{\frac{2}{x}}\geq 3\sqrt[3]{\frac{4}{3}.\frac{1}{9}.\frac{2}{x}}\)

\(\Leftrightarrow \frac{4x}{3}+\frac{2}{3}\sqrt{\frac{2}{x}}\geq 2\)

\(\Rightarrow A\geq 2\)

Vậy \(A_{\min}=2\). Dấu bằng xảy ra khi \(x=\frac{1}{2}\)