Ta có: \(ab+bc+ca=abc\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

Đặt: \(A=\frac{a}{bc\left(a+1\right)}+\frac{b}{ca\left(b+1\right)}+\frac{c}{ab\left(c+1\right)}\)

\(\Rightarrow A=\frac{\frac{1}{b}.\frac{1}{c}}{1+\frac{1}{a}}+\frac{\frac{1}{c}.\frac{1}{a}}{1+\frac{1}{b}}+\frac{\frac{1}{b}.\frac{1}{a}}{1+\frac{1}{c}}\)

Đặt: \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow x+y+z=1\)

\(A=\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\)

Ta có: \(\frac{xy}{z+1}=\frac{xy}{\left(z+x\right)+\left(z+y\right)}\le\frac{1}{4}\left(\frac{xy}{x+z}+\frac{xy}{y+z}\right)\)

Chứng minh tương tự ta được:

\(\frac{yz}{x+1}\le\frac{yz}{x+y}+\frac{yz}{x+z}\)

\(\frac{zx}{y+1}\le\frac{zx}{x+y}+\frac{zx}{y+z}\)

Cộng vế với vế:

\(\Rightarrow A\le\frac{1}{4}\left(x+y+z\right)=\frac{1}{4}\left(đpcm\right)\)

\(a,x=\sqrt{27}-\sqrt{2}\)\(=3\sqrt{3}-\sqrt{2}>3\sqrt{3}-\sqrt{3}=2\sqrt{3}\)

Mà: \(y=\sqrt{3}< 2\sqrt{3}\)

\(\Rightarrow x>y\)

\(b,x=\sqrt{5\sqrt{6}}\Rightarrow x^4=5^2.6=150\)

\(y=\sqrt{6\sqrt{5}}\Rightarrow y^4=6^2.5=180\)

\(\Rightarrow x^4< y^4\Rightarrow x< y\left(x,y>0\right)\)

\(c,x=2m;y=m+2\)

Ta có: \(x-y=2m-\left(m+2\right)=m-2\)

Ta xét các trường hợp:

• Nếu \(m< 2\Rightarrow m-2< 0\Rightarrow x< y\)
• Nếu \(m=2\Rightarrow m-2=0\Rightarrow x=y\)
• Nếu \(m>2\Rightarrow m-2=0\Rightarrow x>y\)

a) ( ac + bd )² + ( ad - bc )² = ( a² + b² ).( c² + d² )

\(\Leftrightarrow\)(ac)² + 2acbd + (bd)² + (ad)² - 2adbc + (bc)² = (ac)² + (ad)² + (bc)² + (bd)²

\(\Leftrightarrow\) 2abcd - 2abcd = 0

\(\Leftrightarrow\) 0 = 0 ( luôn đúng )

b)  ( ac + bd )² \(\le\)( a² + b² ).( c² + d² )

\(\Leftrightarrow\)(ac)² + 2acbd + (bd)² \(\le\)(ac)² + (ad)² + (bc)² + (bd)²

\(\Leftrightarrow\)(ad)² - 2adbc + (bc)² \(\ge\)0

\(\Leftrightarrow\)( ad - bc )² \(\ge\)0 ( luôn đúng )

Ta có: \(A=\left(1+x\right)\left(1+\frac{1}{y}\right)+\left(1+y\right)\left(1+\frac{1}{x}\right)\)

\(=1+\frac{1}{y}+x+\frac{x}{y}+1+\frac{1}{x}+y+\frac{y}{x}\)

\(=\left(x+\frac{1}{2x}\right)+\left(y+\frac{1}{2y}\right)+\left(\frac{y}{x}+\frac{x}{y}\right)+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)+2\)

Lại có: \(x,y\in Z^+\) nên ta có:

• \(x+\frac{1}{2x}\ge\sqrt{2}\)

Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{\sqrt{2}}\)

• \(y+\frac{1}{2y}\ge\sqrt{2}\)

Dấu " = " xảy ra  \(\Leftrightarrow y=\frac{1}{\sqrt{2}}\)

• \(\frac{x}{y}+\frac{y}{x}\ge2\)

Dấu " = " xảy ra \(\Leftrightarrow x=y\)

• \(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\ge\frac{2}{\frac{\sqrt{x^2+y^2}}{2}}=2\sqrt{2}\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=\frac{1}{\sqrt{2}}\)

Từ trên ta suy ra: \(A\ge3\sqrt{2}+4\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=\frac{1}{\sqrt{2}}\)

Vậy \(A_{Min}=3\sqrt{2}+4\)

\(A=\left(x+y\right)+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{1}{x}+\frac{1}{y}\right)+2\ge x+y+\frac{4}{x+y}+4\)

\(\Rightarrow A\ge\left(x+y+\frac{2}{x+y}\right)+\frac{2}{x+y}+4\ge2\sqrt{2}+4+\frac{2}{\sqrt{2\left(x^2+y^2\right)}}=3\sqrt{2}+4\)

Đặt: \(a=\sqrt{2+x};b=\sqrt{2-x}\left(a,b\ge0\right)\)

\(\Rightarrow\hept{\begin{cases}a^2+b^2=4\\a^2-b^2=2x\end{cases}}\)

\(\Rightarrow A=\frac{\sqrt{2+ab}\left(a^3-b^3\right)}{4+ab}=\frac{\sqrt{2+ab}\left(a-b\right)\left(a^2+b^2+ab\right)}{4+ab}\)

\(\Rightarrow A=\frac{\sqrt{2+ab}\left(a-b\right)\left(4+ab\right)}{4+ab}=\sqrt{2+ab}\left(a-b\right)\)

\(\Rightarrow A\sqrt{2}=\sqrt{4+2ab}\left(a-b\right)\)

\(\Rightarrow A\sqrt{2}=\sqrt{\left(a^2+b^2+2ab\right)}\left(a-b\right)=\left(a+b\right)\left(a-b\right)\)

\(\Rightarrow A\sqrt{2}=a^2-b^2=2x\)

\(\Rightarrow A=x\sqrt{2}\)

1

\(x^2-4mx+4m^2-2=0\)

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

\(\Leftrightarrow\left(x-2m+\sqrt{2}\right)\left(x-2m-\sqrt{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2m-\sqrt{2}\\x=2m+\sqrt{2}\end{cases}}\) 

Vậy............