Ta có:

\(\frac{a+1}{1+b^2}=a+1-\frac{\left(a+1\right)b^2}{1+b^2}\ge a+1-\frac{\left(a+1\right)b^2}{2b}=a+1-\frac{ab+b}{2}\left(1\right)\)

Tương tụ ta có:

\(\hept{\begin{cases}\frac{\left(b+1\right)}{1+c^2}\ge b+1-\frac{bc+c}{2}\left(2\right)\\\frac{\left(c+1\right)}{1+a^2}\ge c+1-\frac{ca+a}{2}\left(3\right)\end{cases}}\)

Từ (1), (2), (3) ta có:

\(M\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)

\(=3+3-\frac{ab+bc+ca+3}{2}\)

\(\ge\frac{9}{2}-\frac{\left(a+b+c\right)^2}{6}=3\)

\(ĐK:x\ge3\)

\(\Leftrightarrow5x^2+4=x^2+22x-18+10\sqrt{x.x-6.x+3}\)

\(\Leftrightarrow4x^2-18x+18=10\sqrt{x+3.x^2-6x}=0\)

\(\Leftrightarrow4.x^2-6x+6.x+3-10\sqrt{x+3.x^2-6x}=0\)

\(\Leftrightarrow2\sqrt{x^2-6x}-3\sqrt{x+3}.\sqrt{x^2-6x}-\sqrt{x+3}=0\)

ĐÁP SỐ là mấy  vậy 

Có : 2 > \(\sqrt{3}\) ; 3 > \(\sqrt{4}\) ; ..... ; 1999 > \(\sqrt{2000}\)

=> VT = \(\sqrt{2\sqrt{3\sqrt{4......\sqrt{1999\sqrt{2000}}}}}\)<   \(\sqrt{2\sqrt{3\sqrt{4......\sqrt{1999.1999}}}}\)

= \(\sqrt{2\sqrt{3\sqrt{4.....\sqrt{1999}}}}\) < ........ < \(\sqrt{2\sqrt{3}}\) <  \(\sqrt{2.2}\) = 2

=> ĐPCM

Ta có: \(n=\sqrt{n^2}=\sqrt{1+n^2-1}=\sqrt{1+n-1.n+1}\)

Áp dụng công thức trên với \(n=4,5,6\)ta có:

\(4=\sqrt{1+3.5}=\sqrt{1+3\sqrt{1+4\sqrt{1+5.7}}}=\sqrt{1+3\sqrt{1+\sqrt{4\sqrt{1+...n-1\sqrt{n+1^2}}}}}\)

\(>\sqrt{3\sqrt{4\sqrt{...2000}}}\)

Do đó: \(\sqrt{2+\sqrt{3\sqrt{4\sqrt{...2000}}}}< \sqrt{2+2}=2\)

Ta có:\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=a+b+c\)

\(\Rightarrow\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(a+c\right)}{a+c}+\frac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+a+b+c=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)

kết quả bằng 1

Điều kiện:

\(x,y,z\ge-1\)

Xét các trường hợp, dùng phương pháp dánh giá, CM được: \(x=y=z\) 

Thế vào tìm được nghiệm:

\(x=y=z=\frac{1\pm\sqrt{5}}{x}\)

P/s: Ko chắc