Akai HarumaNguyễn Thành Trương

<= 3/4 nha ko phải a+b+c

minhf lm dc cau nay rồi

Đặt \(\hept{\begin{cases}x=3a+b+c\\y=3b+a+c\\z=3c+a+b\end{cases}\left(x;y;z>0\right)}\)

\(\Rightarrow x+y+z=5a+5b+5c=5\left(a+b+c\right)\)

Lại có: \(a+b+c=x-2a=y-2b=z-2c\)

\(\Rightarrow x+y+z=5\left(x-2a\right)=5\left(y-2b\right)=5\left(z-2c\right)\)

\(\Rightarrow4x-\left(y+z\right)=4\left(3a+b+c\right)-\left(4b+4c+2a\right)=10a\)

Tương tự ta có:\(4y-\left(x+z\right)=10b;4z-\left(x+y\right)=10c\)

\(\Rightarrow10T=\frac{4x-\left(y+z\right)}{x}+\frac{4y-\left(x+z\right)}{y}+\frac{4z-\left(x+y\right)}{z}\)

\(=12-\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}\)

\(=12-\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\right)\)\(\le12-6=6\)(Bđt Cô si)

\(\Rightarrow10T\le6\Rightarrow T\le\frac{6}{10}=\frac{3}{5}\)(Đpcm)

Dấu = khi a=b=c

Xem kỹ lại đề nhé. Anh không nghĩ đề đúng đâu

uk e sorry sửa lại đề rồi đấy 

Chứng minh: \(\frac{5a^3-b^3}{ab+3b^2}\le2a-b\)

Còn lại tương tự

Áp dụng bất đẳng thức Cauchy-Schwartz ta có

\(\frac{ab}{a+3b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right).\)

Tương tự ta có 2 bất đẳng thức khác nữa

\(\frac{bc}{b+3c+2a}=\frac{bc}{\left(b+a\right)+\left(a+c\right)+2c}\le\frac{bc}{9}\left(\frac{1}{b+a}+\frac{1}{a+c}+\frac{1}{2c}\right).\)

\(\frac{ac}{c+3a+2b}=\frac{ac}{\left(a+b\right)+\left(b+a\right)+2a}\le\frac{ac}{9}\left(\frac{1}{c+b}+\frac{1}{b+a}+\frac{1}{2a}\right).\)

Cộng ba bất đẳng thức lại cho ta \(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\)

\(\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)+\frac{bc}{9}\left(\frac{1}{b+a}+\frac{1}{a+c}+\frac{1}{2c}\right)+\frac{ac}{9}\left(\frac{1}{c+b}+\frac{1}{b+a}+\frac{1}{2a}\right)\)

\(=\frac{1}{9}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\frac{1}{9}\left(\frac{ab}{b+c}+\frac{ac}{b+c}\right)+\frac{1}{9}\left(\frac{bc}{a+b}+\frac{ac}{a+b}\right)+\frac{a}{18}+\frac{b}{18}+\frac{c}{18}\)

\(=\frac{a+b+c}{6}.\)   (ĐPCM)

Gọi VT = T

Đặt \(x=3a+b+c;y=3b+c+a;z=3c+a+b\)

\(\Rightarrow x+y+z=5\left(a+b+c\right)=5\left(x-2a\right)=5\left(y-2b\right)\)

\(=5\left(z-2c\right)\)

\(\Rightarrow4x-\left(y+z\right)=10a;4y-\left(z+x\right)=10b;4z-\left(x+y\right)=10c\)

\(\Rightarrow10T=\frac{4x-\left(y+z\right)}{x}+\frac{4y-\left(z+x\right)}{y}+\frac{4z-\left(x+y\right)}{z}\)

\(=12-\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\right)\le12-6=6\)

\(\Rightarrow T\le\frac{6}{10}=\frac{3}{5}\)

Dấu "=" khi a = b = c