Question

[Phanquocvuong]

Cho a,b,c là các số dương .CMR :

T = \(\frac{a}{3a+b+c}+\frac{b}{3b+a+c}+\frac{c}{3c+b+a}\le\frac{3}{5}\)

Answer 1

[Thắng Nguyễn]

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