Lời giải :

\(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)

\(\Leftrightarrow\sqrt{x}=\sqrt{2}+1\)

Thay vào A ta được :

\(A=\frac{3+2\sqrt{2}-2}{2+\sqrt{2}+1}=\frac{1+2\sqrt{2}}{3+\sqrt{2}}\)

Vậy...

=\(\frac{1}{2}.\frac{2\sqrt{3}}{9}-\frac{2\sqrt{3}}{5}+\frac{2\sqrt{3}}{3}\)

=\(\frac{\sqrt{3}}{9}-\frac{2\sqrt{3}}{5}+\frac{2\sqrt{3}}{3}\)

=\(\frac{5\sqrt{3}}{45}-\frac{18\sqrt{3}}{45}+\frac{30\sqrt{3}}{45}\)

=\(\frac{17\sqrt{3}}{45}\)

\(M=\frac{x_1^2+x_2^2+...+x_{2015}^2}{x_1\left(x_2+x_3+...+x_{2015}\right)}\ge\frac{x_1^2+\frac{\left(x_2+x_3+...+x_{2015}\right)^2}{2014}}{x_1\left(x_2+x_3+...+x_{2015}\right)}\)

\(=\frac{x_1}{x_2+x_3+...+x_{2015}}+\frac{x_2+x_3+...+x_{2015}}{2014x_1}\ge2\sqrt{\frac{1}{2014}}=\frac{2}{\sqrt{2014}}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x_2=x_3=...=x_{2015}\\\frac{x_1}{x_2+x_3+...+x_{2015}}=\frac{x_2+x_3+...+x_{2015}}{2014x_1}\end{cases}}\Leftrightarrow x_1=\sqrt{2014}x_2=...=\sqrt{2014}x_{2015}\)

\(6=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\ge\frac{2}{\sqrt[4]{xy}}\)\(\Leftrightarrow\)\(\frac{1}{\sqrt{xy}}\le9\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{9}\)

\(a+b+c+\sqrt{abc}=4\Rightarrow4a+4b+4c+4\sqrt{abc}=16\Rightarrow16-4b-4c=4a+4\sqrt{abc}\)

\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)

\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=2a+\sqrt{abc}\)

Tương tự : \(\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\); \(\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\)

\(\Rightarrow A=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)

Asp dụng bđt AM-GM ta có

\(\frac{\left(\frac{b+c}{a}+1\right)}{2}\ge\sqrt{\frac{b+c}{a}.1}\)

\(\Leftrightarrow\frac{a+b+c}{2a}\ge\sqrt{\frac{b+c}{a}}\) hay  \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)(1)

Tương tự

\(\sqrt{\frac{b}{b+c}}\ge\frac{2b}{a+b+c}\)(2)

\(\sqrt{\frac{c}{c+a}}\ge\frac{2c}{a+b+c}\)(3)

Từ (1),(2),(3)  ta có

\(VT\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)

dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{\frac{a}{a+b}}=1\\\sqrt{\frac{b}{b+c}}=1\\\sqrt{\frac{c}{c+a}}=1\end{cases}}\)(vô lí ) 

Vậy dấu "=" không xảy ra 

do đó \(VT>2\)

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