§1. Bất đẳng thức

Đề thiếu rồi bạn

wut?

\(f\left(x\right)=\left(2-x\right)\left(x+3\right)\le\dfrac{1}{4}\left(2-x+x+3\right)^2=\dfrac{25}{4}\)

\(f\left(x\right)_{max}=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)

Sửa: Cho a+b<0

\(BĐT\Leftrightarrow\dfrac{\left(a+b\right)^3}{8}\ge\dfrac{a^3+b^3}{2}\\ \Leftrightarrow2\left(a+b\right)^3\ge8\left(a^3+b^3\right)\\ \Leftrightarrow2\left(a^3+b^3\right)+6ab\left(a+b\right)\ge8\left(a^3+b^3\right)\\ \Leftrightarrow6ab\left(a+b\right)-6\left(a^3+b^3\right)\ge0\\ \Leftrightarrow6\left[ab\left(a+b\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\right]\ge0\\ \Leftrightarrow6\left(a+b\right)\left(-a^2+2ab-b^2\right)\ge0\\ \Leftrightarrow-6\left(a+b\right)\left(a-b\right)^2\ge0\left(\text{luôn đúng do }-6< 0;a+b< 0\right)\)

Dấu \("="\Leftrightarrow a=b< 0\)

Áp dụng BĐT cosi:

\(\left(a+b+b+c+c+a\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\\ \ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\dfrac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\\ \Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\\ \Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(đpcm\right)\)

Dấu \("="\Leftrightarrow a=b=c\)

Áp dụng BĐT cosi:

\(\left(2+a+b\right)\left(a+4b+ab\right)\ge3\sqrt[3]{2ab}\cdot3\sqrt[3]{4a^2b^2}=9\sqrt[3]{8a^3b^3}=9\cdot2ab=18ab\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a=b=2\\a=4b=ab\end{matrix}\right.\left(\text{vô lí}\right)\)

Vậy dấu \("="\) ko xảy ra hay \(\left(2+a+b\right)\left(a+4b+ab\right)>18ab\)

\(A\ge2\cdot3=6\)

Dấu '=' xảy ra khi x=1

mjk cx đag tìm bài này, giải giúp mình với