Question

[Nguyễn Minh Quý]

1) (a+b+c)(1/(a+b)+1/(b+c)+1/(c+a))>=9/2

2) (a+b)(a^2+b^2)/4<=(a^3+b^3)/2

3) a/(2a+b)+b/(2b+a)<=2/3

4) (a^2+2)(b^2+2)(c^2+2)>=abc16√2

Hộ mình nhanh nha

Answer 1

[Phùng Minh Quân]

Thiếu \(a,b\ge0\) nhé 

\(1)\) Cauchy-Schwarz dạng Engel : 

\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{9\left(a+b+c\right)}{2\left(a+b+c\right)}=\frac{9}{2}\) ( đpcm ) 

\(2)\)

\(\frac{\left(a+b\right)\left(a^2+b^2\right)}{4}=\frac{a^3+b^3+ab^2+a^2b}{4}=\frac{a^3+b^3+ab\left(a+b\right)}{4}\)

Cần CM : \(a^3+b^3\ge ab\left(a+b\right)\)

\(\Leftrightarrow\)\(\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)

\(\Leftrightarrow\)\(\left(a+b\right)\left(a^2-ab+b^2-ab\right)=\left(a+b\right)\left(a-b\right)^2\ge0\) ( đúng ) 

\(\frac{a^3+b^3+ab\left(a+b\right)}{4}=\frac{2\left(a^3+b^3\right)}{4}=\frac{a^3+b^3}{2}\) ( đpcm ) 

3,4 làm sau