Question

Answer 1

Ta có: \(P=3x+\dfrac{4}{x^2}\ge3.2+\dfrac{4}{2^2}=6+1=7\)(do \(x\ge2\))

\(\Rightarrow Min_P=7\)

Dấu "=" xảy ra khi \(x=2\)

Vậy \(Min_P=7\) khi \(x=2\)

Answer 2

Do \(x\ge2\) nên \(3x;\dfrac{4}{x^2}>0\)

Áp dụng BĐT Cauchy cho \(2\) số dương là \(3x\) và \(\dfrac{4}{x^2}\) ta được:

\(3x+\dfrac{4}{x^2}\ge2\sqrt{3x.\dfrac{4}{x^2}}=2\sqrt{\dfrac{12}{x}}\)

Dấu "=" xảy ra khi \(3x=\dfrac{4}{x^2}\Leftrightarrow x^3=\dfrac{4}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{4}{3}}\)

Khi đó \(2\sqrt{\dfrac{12}{x}}=2\sqrt{\dfrac{12}{\sqrt[3]{\dfrac{4}{3}}}}=2\sqrt{6\sqrt[3]{6}}\)

\(\Rightarrow Min_P=2\sqrt{6\sqrt[3]{6}}\)

Vậy \(Min_P=2\sqrt{6\sqrt[3]{6}}\) khi \(x=\sqrt[3]{\dfrac{4}{3}}\)