Áp dụng bđt Cô si cho 3 số ta đc

\(\frac{x}{y^2+2}+\frac{y}{z^2+2}+\frac{z}{x^2+2}\ge3\sqrt[3]{\frac{xyz}{\left(y^2+2\right)\left(z^2+2\right)\left(x^2+2\right)}}\)

\(VT\ge3\sqrt[3]{\frac{1}{27}=}1\)

Dấu " = " xảy ra <=> x = y = z = 1

p/s : quên cách làm khúc giữa

Áp dụng bất đẳng thức Cô si cho 3 số thực ko âm ta đc :

\(\frac{x}{y^2+2}+\frac{y}{z^2+2}+\frac{z}{x^2+2}\ge3\sqrt[3]{\frac{xyz}{\left(y^2+2\right)\left(z^2+2\right)\left(x^2+2\right)}}\)

\(\Rightarrow VT\ge3\sqrt[3]{\frac{1}{1+2y^2x^2+2z^2x^2+2z^2y^2+4x^2+4z^2+4y^2+8}}\)( phân tích đa thức thành nhân tử )

\(\Rightarrow VT\ge3\sqrt[3]{\frac{1}{9+\frac{2}{z^2}+\frac{2}{y^2}+\frac{2}{x^2}+4x^2+4z^2+4y^2}}\)( vì \(xyz=1\Rightarrow x^2y^2z^2=1\Rightarrow x^2y^2=\frac{1}{z^2}\)các phân số khác chứng minh tương tự )

\(\Rightarrow VT\ge3\sqrt[3]{\frac{1}{9+\frac{2+4z^4}{z^2}+\frac{2+4y^4}{y^2}+\frac{2+4x^4}{x^2}}}\)( quy đồng mẫu số  ) ( A )

Áp dụng bất đẳng thức Cô si cho 3 số thực ko âm ta được :

\(\frac{2+4z^4}{z^2}+\frac{2+4y^4}{y^2}+\frac{2+4x^4}{x^2}\ge3\sqrt[3]{\frac{\left(2+4z^4\right)\left(2+4y^4\right)\left(2+4x^4\right)}{x^2y^2z^2}}\) ( 1 )

Ta có :

\(2+4x^4\ge2+4.1^4=6\) ( 2 ) 

\(2+4y^4\ge2+4.1^4=6\) ( vì x^4 , y^4 , z^4 đều là các lũy thừa số mũ chẵn ) ( 3 )

\(2+4z^4\ge2+4.1^4=6\)( 4 ) 

x^2 . y^2 . z^2 = ( xyz )^2 = 1^2 = 1 ( 5 )

Từ ( 1 ) , ( 2 ) , ( 3 ) , ( 4 ) , ( 5 )  suy ra :

\(\frac{2+4z^4}{z^2}+\frac{2+4y^4}{y^2}+\frac{2+4x^4}{x^2}\ge3\sqrt[3]{\frac{6^3}{1}}=18\) ( B )

Thay B vào A ta đc  :

\(\Rightarrow VT\ge3\sqrt[3]{\frac{1}{9+\frac{2+4z^4}{z^2}+\frac{2+4y^4}{y^2}+\frac{2+4x^4}{x^2}}}\ge3\sqrt[3]{\frac{1}{9+18}}=1\)

a.ĐK:\(x\ge0;x\ne1\)

\(M=\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\)

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

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

b.\(\frac{3\sqrt{x}}{x-\sqrt{x}+1}< 1\Leftrightarrow3\sqrt{x}< x+\sqrt{x}+1\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)

(Luôn đúng với mọi \(x\ge0;x\ne1\)) . Ta có đpcm

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\)\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

\(\sqrt{4x+1}-\sqrt{3x+1}=1\)

\(\Rightarrow\left(\sqrt{4x+1}-\sqrt{3x+1}\right)^2=1\)

\(\Rightarrow|4x+1|-2\sqrt{\left(4x+1\right)\left(3x+1\right)}+|3x+1|=1\)

Dùng bảng xét dấu mà xét nốt bạn nhé

 Điều kiện x;y >=1Ta có: \(\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}\ge\frac{1}{1+xy}\Leftrightarrow\frac{2}{\left(1+x\right)^2}+\frac{2}{\left(1+y\right)^2}\ge\frac{2}{1+xy}\)

Ta có: \(\hept{\begin{cases}\left(1+x\right)^2\le\left(1^2+1^2\right)\left(x^2+1^2\right)=2\left(x^2+1\right)\\\left(1+y\right)^2\le2\left(y^2+1\right)\end{cases}}\)

Cần cm: \(\frac{1}{x^2+1}+\frac{1}{y^2+1}\ge\frac{2}{1+xy}\)

\(\Leftrightarrow\frac{x^2+y^2+2}{\left(x^2+1\right)\left(y^2+1\right)}\ge\frac{2}{1+xy}\)

\(\Leftrightarrow\left(x^2+y^2+2\right)\left(1+xy\right)\ge2\left(x^2+1\right)\left(y^2+1\right)\)

\(\Leftrightarrow x^2+x^3y+y^2+y^3x+2+2xy\ge2x^2y^2+2x^2+2y^2+2\)

\(\Leftrightarrow x^3y+xy^3+2xy-x^2-y^2-2x^2y^2\ge0\)

\(\Leftrightarrow xy\left(x^2+y^2-2xy\right)-\left(x^2-2xy+y^2\right)=\left(xy-1\right)\left(x-y\right)^2\ge0\)(đúng)

"=" khi x=y=1

Đề sai thì  phải ah.

Với \(x=1;y=2\) ta có:

\(S=\frac{1}{\left(1+1\right)^2}+\frac{1}{\left(1+2\right)^2}\ge\frac{1}{1+1\cdot2}\)

\(S=\frac{1}{4}+\frac{1}{9}\ge\frac{1}{3}\)

\(S=\frac{13}{36}\ge\frac{1}{3}\left(VL\right)\)