Cho hàm số f(x) có đạo hàm dương, liên tục trên đoạn [0;1]

* Hướng dẫn giải

Đáp án D

Giả thiết $\iff 3\underset{0}{\overset{1}{\int }}{\left[\sqrt{\mathrm{f}\text{'}\left(\mathrm{x}\right)}.\mathrm{f}\left(\mathrm{x}\right)\right]}^2\mathrm{dx}\frac{1}{3}\le 2\underset{0}{\overset{1}{\int }}\sqrt{\mathrm{f}\text{'}\left(\mathrm{x}\right)}.\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

$\begin{array}{l}\iff \underset{0}{\overset{1}{\int }}{\left[3\sqrt{\mathrm{f}\text{'}\left(\mathrm{x}\right)}.\mathrm{f}\left(\mathrm{x}\right)\right]}^2\mathrm{dx}-2\underset{0}{\overset{1}{\int }}3\sqrt{\mathrm{f}\text{'}\left(\mathrm{x}\right)}.\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\underset{0}{\overset{1}{\int }}\mathrm{dx}\le 0\\ \iff \underset{0}{\overset{1}{\int }}{\left[3\sqrt{\mathrm{f}\text{'}\left(\mathrm{x}\right)}.\mathrm{f}\left(\mathrm{x}\right)-1\right]}^2\mathrm{dx}\le 0\end{array}$