CMR \(\dfrac{1}{a^3+1}+\dfrac{1}{b^3+1}+\dfrac{1}{c^3+1}\le\dfrac{1}{ab^2+1}+\dfrac{1}{bc^2+1}+\dfrac{1}{ca^2+1}\)

\(\dfrac{a+b}{c^2+ab}+\dfrac{b+c}{a^2+bc}+\dfrac{c+a}{b^2+ac}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

Cho abc=1

CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\)

\(\left(x+1\right)\sqrt{x+2}+\left(x+6\right)\sqrt{x+7}\ge x^2+7x+12\)

Bài 7) 

\(bđt\Leftrightarrow4\left(a+b+c\right)\left(a^2+b^2+c^2\right)-3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3\)

\(\Leftrightarrow a^3+b^3+c^3+4ab\left(a+b\right)+4bc\left(b+c\right)+4ac\left(a+c\right)\ge\left(a+b+c\right)^3\)

\(\Leftrightarrow4ab\left(a+b\right)+4bc\left(b+c\right)+4ac\left(a+c\right)\ge3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)+6abc\)\(\Leftrightarrow ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\ge6abc\)

\(\Leftrightarrow\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)

(Đúng theo Cô Si)

"=" khi a=b=c=1

chi gioi qua 

\(\left(x;2y;3z\right)\Rightarrow\left(a;b;c\right)\)

\(\left\{{}\begin{matrix}a+b+c=18\\CM:\dfrac{b+c+5}{1+a}+\dfrac{c+a+5}{1+b}+\dfrac{a+b+5}{1+c}\ge\end{matrix}\right.\dfrac{51}{7}\)

\(bdt\Leftrightarrow\dfrac{23-a}{a+1}+\dfrac{23-b}{b+1}+\dfrac{23-c}{c+1}\ge\dfrac{51}{7}\)

\(\Leftrightarrow\dfrac{24}{a+1}-1+\dfrac{24}{b+1}-1+\dfrac{24}{c+1}-1\ge\dfrac{51}{7}\)

\(\Leftrightarrow\dfrac{24}{a+1}+\dfrac{24}{b+1}+\dfrac{24}{c+1}\ge\dfrac{72}{7}\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge\dfrac{3}{7}\)

Mà Theo Cauchy-Schwarz thì: \(VT\ge\dfrac{9}{a+b+c+3}=\dfrac{9}{21}=\dfrac{3}{7}\)

"=".....

Câu 266 là >= chứ nhỉ?