theo bài ra ta có 

\(\frac{1}{x}+\frac{1}{2y}+\frac{1}{3z}=0\Leftrightarrow6yz+3xz+2xy=0\)       (1)

\(x+2y+3z=4\Leftrightarrow\left(x+2y+3z\right)^2=16\)

                                       \(\Leftrightarrow x^2+4y^2+9z^2+2\left(6yz+3xz+2xy\right)=16\)(2)

                               thay  (1) vào (2)  ta được 

\(x^2+4y^2+9z^2=16\)

Đặt \(x=a;2y=b;3z=c\Rightarrow a+b+c=3\) 

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

Áp dụng Bđt Cô si ngược dấu ta có:

\(T=\text{∑}a-\frac{a^2b}{1+b^2}\ge\text{∑}a-\frac{a^2b}{2b}=\text{∑}a-\frac{ab}{2}\)

\(=a+b+c-\frac{ab+bc+ca}{2}\ge a+b+c-\frac{\left(ab+bc+ca\right)^2}{6}\)\(=3-\frac{3^2}{6}=\frac{3}{2}\)

Dấu = khi \(a=b=c=1\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\\z=\frac{1}{3}\end{cases}}\)

Lời giải:

Vì $x,y,z$ tỉ lệ với $5,4,3$ nên:

$\frac{x}{5}=\frac{y}{4}=\frac{z}{3}$

Đặt $\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=k\Rightarrow x=5k; y=4k; z=3k$.

Khi đó:

$P=\frac{x+2y-3z}{x-2y+3z}=\frac{5k+2.4k-3.3k}{5k-2.4k+3.3k}$

$=\frac{5k+8k-9k}{5k-8k+9k}=\frac{4k}{6k}=\frac{2}{3}$

vì \(\left(x+2y\right)^2+\left(y-1\right)^2+\left(x-z\right)^2=0\Leftrightarrow\)

       \(\left(x+2y\right)^2=0\Leftrightarrow x+2y=0\Leftrightarrow x=2y\left(1\right)\)

       \(\left(y-1\right)^2=0\Leftrightarrow y-1=0\Leftrightarrow y=1\left(2\right)\)

          \(\left(x-z\right)^2=0\Leftrightarrow x-z=0\Leftrightarrow x=z\left(3\right)\)

 \(\left(1\right)\left(2\right)\left(3\right)\Rightarrow2y=x=y=2\left(4\right)\)

                      \(\left(4\right)\Leftrightarrow A=2+2+3\times2=10\)