\(\sqrt{4x+2\sqrt{x}+1}\le\sqrt{4x+\dfrac{1}{2}\left(2^2+x\right)+1}=\sqrt{\dfrac{9x}{2}+3}\)

\(=\dfrac{1}{\sqrt{21}}.\sqrt{21}.\sqrt{\dfrac{9x}{2}+3}\le\dfrac{1}{2\sqrt{21}}\left(21+\dfrac{9x}{2}+3\right)=\dfrac{1}{2\sqrt{21}}\left(\dfrac{9x}{2}+24\right)\)

Tương tự và cộng lại:

\(A\le\dfrac{1}{2\sqrt{21}}\left(\dfrac{9}{2}\left(x+y+z\right)+72\right)=3\sqrt{21}\)

\(A_{max}=3\sqrt{21}\) khi \(x=y=z=4\)

\(A=1\sqrt{4x+2\sqrt{x}+1}+1.\sqrt{4y+2\sqrt{y}+1}+1\sqrt{4z+2\sqrt{z}+1}\)

\(\le\sqrt{\left(1+1+1\right)\left(4\left(x+y+z\right)+2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3\right)}\)

\(=\sqrt{3.\left[51+\dfrac{4\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{2}\right]}\)

\(\le\sqrt{3.\left[51+\dfrac{x+y+z+12}{2}\right]}\)

\(=\sqrt{189}\)

Dấu "=" xảy ra <=> x = y = z = 4

Các biểu thức ở trong căn đều đưa được về bình phương
\(\sqrt{4x+2\sqrt{x}+1}=\sqrt{\left(2\sqrt{x}+1\right)^2}=\left|2\sqrt{x}+1\right|=2\sqrt{x}+1\)

Tương tự với hai căn còn lại ta sẽ có biểu thức đề cho tương đương với
\(2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3\)

\(A=\sum\sqrt{4x+2\sqrt{x}+1}\)

\(Max_A=+\infty\)

\("="x=y=z=+\infty\)

ta có \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\Rightarrow x+y+z\le3\)

ta có :\(\sqrt{4x+5}=\frac{\sqrt{9\left(4x+5\right)}}{3}\le\frac{9+4x+5}{2\times3}=\frac{2x+7}{3}\)

tương tự ta sẽ có  ; \(A\le\frac{2x+7}{3}+\frac{2y+7}{3}+\frac{2z+7}{3}=\frac{2}{3}\left(x+y+z\right)+7\le\frac{2}{3}\times3+7=9\)

Vậy GTLN của A=9

dấu bằng xảy ra khi x= y= z =1

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2\left(xy+yz+zx\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=3.3=9\)

\(\Rightarrow x+y+z\le3\).

\(A=\sqrt{4x+5}+\sqrt{4y+5}+\sqrt{4z+5}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(4x+5+4y+5+4z+5\right)}\)

\(=\sqrt{3\left[4\left(x+y+z\right)+15\right]}=9\)

Dấu \(=\)khi \(x=y=z=1\).

Áp dụng bất đẳng thức Bunyakovsky:

\(NL^2=\left(\sqrt{4x+2\sqrt{x}+1}+\sqrt{4y+2\sqrt{y}+1}+\sqrt{4z+2\sqrt{z}+1}\right)^2\)

\(\le\left(1^2+1^2+1^2\right)\left(4x+2\sqrt{x}+1+4y+2\sqrt{y}+1+4z+2\sqrt{z}+1\right)\)

\(=3\left(4x+4y+4z\right)+3\left(2\sqrt{x}+2\sqrt{y}+2\sqrt{z}\right)+3\left(1+1+1\right)\)

\(=12\left(x+y+z\right)+6\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+9\)

\(=153+6\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

Mặt khác,theo Bunyakovsky: \(\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\le3\left(x+y+z\right)=36\)

\(\Rightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}\le6\)

\(\Rightarrow153+6\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\le153+36=189\)

\(\Rightarrow NL\le\sqrt{189}\)

Dấu "=" xảy ra khi: \(x=y=z=4\)

không biết :))))