A = T; G= X 

mà A + T + G + X = 4800 => A + G = 2400 

Theo bài ra : G - A = 600 

=> G = ( 2400 + 600) : 2 = 1500 => X = G = 1500

A = 2400 - 1500 = 900 => T = A = 900

Theo giả thiết, ta có: \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=2019\Rightarrow x+y+z=2019xyz\)

Xét \(\sqrt{2019x^2+1}=\sqrt{\frac{x^2+xy+xz}{yz}+1}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{yz}}\le\frac{\frac{x+y}{y}+\frac{x+z}{z}}{2}=\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)+1\)\(\Rightarrow\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le\frac{x^2+1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)+1}{x}=x+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)+\frac{2}{x}\)

Tương tự, ta có: \(\frac{y^2+1+\sqrt{2019y^2+1}}{y}\le y+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)+\frac{2}{y}\); \(\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le z+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{2}{z}\)

Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{x^2+1+\sqrt{2019x^2+1}}{x}+\frac{y^2+1+\sqrt{2019y^2+1}}{y}+\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le\left(x+y+z\right)+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=2019xyz+3.\frac{xy+yz+zx}{xyz}\le2019xyz+3.\frac{\frac{\left(x+y+z\right)^2}{3}}{xyz}=2019xyz+\frac{\left(x+y+z\right)^2}{xyz}=2019xyz+2019^2xyz=2019.2020xyz\)Đẳng thức xảy ra khi \(x=y=z=\frac{1}{\sqrt{673}}\)

Vì a,b,c > 0 nên \(\frac{a}{a+b}< 1\Rightarrow\frac{a}{a+b}< \frac{a+c}{a+b+c}\)

Tương tự: \(\frac{b}{b+c}< \frac{b+a}{a+b+c};\frac{c}{c+a}< \frac{c+b}{a+b+c}\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< 2\left(1\right)\)

Ta có: \(\sqrt{\frac{a}{b+c}}=\frac{a}{\sqrt{a\left(b+c\right)}}\)

Vì a,b,c>0 nên theo BĐT Cô-si ta có:

\(\frac{a+\left(b+c\right)}{2}\ge\sqrt{a\left(b+c\right)}>0\)

\(\Leftrightarrow\frac{2}{a+b+c}\le\frac{1}{\sqrt{a\left(b+c\right)}}\)

\(\Leftrightarrow\frac{2a}{a+b+c}\le\frac{a}{a\left(b+c\right)}\Leftrightarrow\frac{2a}{a+b+c}\le\sqrt{\frac{a}{b+c}}\)

Tương tự: \(\frac{2b}{a+b+c}\le\sqrt{\frac{b}{a+c}};\frac{2c}{a+b+c}\le\sqrt{\frac{c}{b+a}}\)

\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge2\)

Dấu '=' xảy ra khi a=b+c;b=c+a;c=a+b

tức là a=b=c(vô lý)

\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2\left(2\right)\)

Từ 1,2 => đpcm

\(\left(\frac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}+\frac{5}{\sqrt{5}}\right):\frac{1}{\sqrt{5}-\sqrt{3}}=\left(\frac{\sqrt{3}.\left(1+\sqrt{5}\right)}{1+\sqrt{5}}+\sqrt{5}\right).\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(\sqrt{5}+\sqrt{3}\right).\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)