Ta có: \(abc\le\frac{\left(a+b+c\right)^3}{27}\)  ; \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)

Mà \(a^2+b^2+c^2=3abc\)

=>\(\frac{\left(a+b+c\right)^2}{3}\le\frac{\left(a+b+c\right)^3}{27}.3\)

=> \(a+b+c\ge3\)

Áp dụng bđt bunhia dạng phân thức ta có:

\(M\ge\frac{\left(a+b+c\right)^2}{a+b+c+6}\)

Đặt \(a+b+c=x\left(x\ge3\right)\)

=> \(M\ge\frac{x^2}{x+6}\)

Xét \(\frac{x^2}{x+6}\ge\frac{5}{9}x-\frac{2}{3}\)

<=>\(x^2\ge\frac{5}{9}x^2+\frac{8}{3}x-4\)

<=>\(\left(\frac{2}{3}x-2\right)^2\ge0\)(luôn đúng)

=> \(M\ge\frac{5}{9}x-\frac{2}{3}\ge\frac{5}{9}.3-\frac{2}{3}=1\)

=>\(MinM=1\)xảy ra khi a=b=c=1

\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

\(=\sqrt{5}\left(2-5-4+11\right)=4\sqrt{5}\)

\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

\(=2\sqrt{5}-\sqrt{5.25}-\sqrt{5.16}+\sqrt{5.121}\)

\(=2\sqrt{5}-\sqrt{5}.\sqrt{25}-\sqrt{5}.\sqrt{16}+\sqrt{5}.\sqrt{121}\)

\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

\(=4\sqrt{5}\)

Ta có x + y + z = 2

=> (x + y + z)² = 4

<=> x² + y² + z² + 2(xy + yz + zx) = 4

<=> 2 + 2(xy + yz + zx) = 4

<=> xy + yz + zx = 1

<=> \(\frac{xyz}{x}+\frac{xyz}{y}+\frac{xyz}{z}=1\)

<=>  \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\left(\text{đpcm}\right)\)

ĐK \(-2\ge x\le2\)

Ta có \(9x^2+2\sqrt{x^2-4}=36\)

\(\Leftrightarrow9\left(x^2-4\right)+2\sqrt{x^2-4}=0\)

Đặt \(\sqrt{x^2-4}=t\left(t\ge0\right)\Rightarrow x^2-4=t^2\)ta có 

\(9t^2+2t=0\Leftrightarrow t\left(9t+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=0\\9t+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=0\left(TM\text{Đ}K\right)\\t=-\frac{2}{9}\left(lo\text{ại}\right)\end{cases}}}\)

\(\Leftrightarrow\sqrt{x^2-4}=0\Leftrightarrow x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\left(TM\text{Đ}K\right)\)

điều kiện \(x\ge0;P\ge0\)

Để chứng minh \(p>\sqrt{P}\)luôn đúng ta cần chứng minh P>1 luôn đúng.

Giả sử P>1 \(\Leftrightarrow\)\(\frac{x+16}{\sqrt{x}+3}>1\)\(\Leftrightarrow\)\(x+16>\sqrt{x}+3\)\(\Leftrightarrow\)\(x-\sqrt{x}+13>0\)

\(\Leftrightarrow\)\(x+\sqrt{x}+\frac{1}{4}+12,75>0\)\(\Leftrightarrow\)\(\left(\sqrt{x}+\frac{1}{2}\right)^2+12,75>0\)luôn luôn đúng

như vậy P luôn luôn >1 là đúng\(\Leftrightarrow\)\(p>\sqrt{P}\)luôn đúng (đpcm)