Quay lại đáp án

Cách 3:Đặt $\begin{cases}x=3a+b+c \\ y=3b+a+c \\ z=3c+b+a\end{cases}$$\Rightarrow x+y+z=5(a+b+c)=5(x-2a)=5(y-2b)=5(z-2c)$$\Rightarrow \begin{cases}4x-(y+z)=10a \\ 4y-(x+z)=10b \\ 4z-(y+x)=10c \end{cases}$$\Rightarrow 10T=\sum \frac{4x-(y+z)}{x}=12-(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z})$Theo Cauchy $\Rightarrow 10T\leq 12-6=6$$\Rightarrow T\leq \frac{3}{5}$Đăngt thức xảy ra khi $a=b=c./$

Cách 3:Đặt $\begin{cases}x=3a+b+c \\ y=3b+a+c \\ z=3c+b+a\end{cases}$$\Rightarrow x+y+z=5(a+b+c)=5(x-2a)=5(y-2b)=5(z-2c)$$\Rightarrow \begin{cases}4x-(y+z)=10a \\ 4y-(x+z)=10b \\ 4z-(y+x)=10c \end{cases}$$\Rightarrow 10T=\sum \frac{4x-(y+z)}{x}=12-(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z})$Theo Cauchy $\Rightarrow 10T\leq 12-6=6$$\Rightarrow T\leq \frac{3}{5}$Đẳng thức xảy ra khi $a=b=c./$

Cách 3:Đặt $\begin{cases}x=3a+b+c \\ y=3b+a+c \end{cases}$ và $z=3c+b+a$$\Rightarrow x+y+z=5(a+b+c)=5(x-2a)=5(y-2b)=5(z-2c)$$\Rightarrow \begin{cases}4x-(y+z)=10a \\ 4y-(x+z)=10b \end{cases}$và: $4z-(y+x)=10c$$\Rightarrow 10T=\Sigma \frac{4x-(y+z)}{x}=12-(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z})$Theo Cauchy $\Rightarrow 10T\leq 12-6=6$$\Rightarrow T\leq \frac{3}{5}$Đăngt thức xảy ra khi $a=b=c./$

Cách 3:Đặt $\begin{cases}x=3a+b+c \\ y=3b+a+c \\ z=3c+b+a\end{cases}$$\Rightarrow x+y+z=5(a+b+c)=5(x-2a)=5(y-2b)=5(z-2c)$$\Rightarrow \begin{cases}4x-(y+z)=10a \\ 4y-(x+z)=10b \\ 4z-(y+x)=10c \end{cases}$$\Rightarrow 10T=\sum \frac{4x-(y+z)}{x}=12-(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z})$Theo Cauchy $\Rightarrow 10T\leq 12-6=6$$\Rightarrow T\leq \frac{3}{5}$Đăngt thức xảy ra khi $a=b=c./$