Đặt \(P=\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Do x,y,z là các số thực dương nên ta biến đổi \(P=\frac{1}{\sqrt{1+\frac{1}{x^2}}}+\frac{1}{\sqrt{1+\frac{1}{y^2}}}+\frac{1}{\sqrt{1+\frac{1}{z^2}}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

Đặt \(a=\frac{1}{x^2};b=\frac{1}{y^2};c=\frac{1}{z^2}\left(a,b,c>0\right)\)thì \(xy+yz+zx=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}=1\)và \(P=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}+a+b+c\)

Biến đổi biểu thức P=\(\left(\frac{1}{2\sqrt{a+1}}+\frac{1}{2\sqrt{a+1}}+\frac{a+1}{16}\right)+\left(\frac{1}{2\sqrt{b+1}}+\frac{1}{2\sqrt{b+1}}+\frac{b+1}{16}\right)\)\(+\left(\frac{1}{2\sqrt{c+1}}+\frac{1}{2\sqrt{c+1}}+\frac{c+1}{16}\right)+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{b}-\frac{3}{16}\)

Áp dụng Bất Đẳng Thức Cauchy ta có

\(P\ge3\sqrt[3]{\frac{a+1}{64\left(a+1\right)}}+3\sqrt[3]{\frac{b+1}{64\left(b+1\right)}}+3\sqrt[3]{\frac{c+1}{64\left(c+1\right)}}+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{16}-\frac{3}{16}\)

\(=\frac{33}{16}+\frac{15}{16}\left(a+b+c\right)\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{abc}\)

Mặt khác ta có \(1=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge3\sqrt[3]{\frac{1}{abc}}\Leftrightarrow abc\ge27\)

\(\Rightarrow P\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{27}=\frac{33}{16}+\frac{15}{16}\cdot9=\frac{21}{2}\)

Dấu "=" xảy ra khi a=b=c hay \(x=y=z=\frac{\sqrt{3}}{3}\)

a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)

\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)

\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)

\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)

\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)

\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)

\(=\frac{3x^2-3y^2}{50}\)

c) \(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)

\(=\frac{2}{xy}:\frac{y-x}{xy}-\frac{\left(x+y\right)\left(x-y\right)}{\left(x-y\right)^2}\)

\(=\frac{2}{y-x}-\frac{x+y}{x-y}\)

\(=\frac{2}{y-x}+\frac{x+y}{y-x}\)

\(=\frac{x+y+2}{y-x}\)

d) \(\frac{\frac{x-y}{x+y}-\frac{x+y}{x-y}}{1-\frac{x^2}{x^2+y^2}}\)

\(=\frac{\frac{x^2-2xy+y^2}{x^2-y^2}-\frac{x^2+2xy+y^2}{x^2-y^2}}{\frac{y^2}{x^2+y^2}}\)

\(=\frac{\frac{2x^2+2y^2}{x^2-y^2}}{\frac{y^2}{x^2+y^2}}\)

\(=\frac{2x^2+2y^2}{x^2-y^2}.\frac{x^2+y^2}{y^2}\)

\(=\frac{2x^4+4x^2y^2+2y^4}{x^2y^2-y^4}\)

\(VT=\Sigma_{cyc}\frac{2\sqrt{x}}{x^3+y^2}\le\Sigma_{cyc}\frac{2\sqrt{x}}{2\sqrt{x^3y^2}}=\Sigma_{cyc}\frac{1}{\sqrt{x^2y^2}}=\Sigma_{cyc}\frac{1}{xy}\)

\(=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\) (áp dụng BĐT quen thuộc \(ab+bc+ca\le a^2+b^2+c^2\))

Đẳng thức xảy ra khi x = y = z = 1

Sửa đề : \(\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\hept{\begin{cases}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\\\frac{2\sqrt{y}}{y^3+z^2}\le\frac{2\sqrt{y}}{2yz\sqrt{y}}=\frac{1}{yz}\\\frac{2\sqrt{z}}{z^3+x^2}\le\frac{2\sqrt{z}}{2xz\sqrt{z}}=\frac{1}{xz}\end{cases}}\)

\(\Rightarrow VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(1\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2y^2}}=\frac{2}{xy}\\\frac{1}{y^2}+\frac{1}{z^2}\ge2\sqrt{\frac{1}{y^2z^2}}=\frac{2}{yz}\\\frac{1}{z^2}+\frac{1}{x^2}\ge2\sqrt{\frac{1}{x^2z^2}}=\frac{2}{xz}\end{cases}}\)

\(\Rightarrow2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\ge2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(2\right)\)

Từ (1) và (2) :

\(\Rightarrow VT\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

\(\Leftrightarrow\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\left(đpcm\right)\)

Chúc bạn học tốt !!!

\(\hept{\begin{cases}y=\frac{x^2+\frac{1}{x^2}}{x^2-\frac{1}{x^2}}=\frac{x^4+1}{x^4-1}=a\\z=\frac{x^4+\frac{1}{x^4}}{x^4-\frac{1}{x^4}}=\frac{x^8+1}{x^8-1}\end{cases}}\)

\(\Rightarrow x^4=\frac{y+1}{y-1}\)

Thế vô z được

\(z=\frac{\left(\frac{y+1}{y-1}\right)^2+1}{\left(\frac{y+1}{y-1}\right)-1}=\frac{y^2+1}{2y}\)

Giờ thì thế \(y=\sqrt{2}+\sqrt{3}\)vô đi