Cô-si Engel :

\(P=\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+b+c+6}=\frac{a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{a+b+c+6}\)

\(\ge\frac{a+b+c+2.3\sqrt[3]{\sqrt{ab}.\sqrt{bc}.\sqrt{ac}}}{a+b+c+6}=\frac{a+b+c+6\sqrt[3]{abc}}{a+b+c+6}=\frac{a+b+c+6}{a+b+c+6}=1\)

Nguyễn Linh Chi Thanks cô,e đổi biến lộn ạ:(

Đặt \(a=\frac{x}{y};b=\frac{y}{z};c=\frac{z}{x}\)

Ta có:

\(P=\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\)

\(=\frac{1}{1+\frac{2}{a}}+\frac{1}{1+\frac{2}{b}}+\frac{1}{1+\frac{2}{c}}\)

\(=\frac{1}{1+\frac{2y}{x}}+\frac{1}{1+\frac{2z}{y}}+\frac{1}{1+\frac{2x}{z}}\)

\(=\frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}\)

\(=\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2zx}\)

\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)

Dấu "=" xảy ra tại \(a=b=c=1\)

Theo Bunhiacopski ta luôn có:

\(\left(x-y\right)^2=\left[1\cdot x+\left(-\frac{1}{2}\right)\cdot2y\right]^2\le\left(1^2+\frac{1}{4}\right)\left(x^2+4y^2\right)=\frac{5}{2}\)

\(\Rightarrow\left|x-y\right|\le\frac{\sqrt{5}}{2}\left(đpcm\right)\)

47,15

29,9

19,86

3,34

15,25

9,9

TRL:

34,5+12,65 = 47,15

21+8,9 = 29,9

7,86+12 = 29,86

6,4-3,14 = 3,26

25-9,75 = 14,25

18,9-9 =9,9

#HuyềnAnh#