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:

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

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

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

Cách 1: 3√1728 : 3√64 = 12:4 = 3

Cách 2: 3√1728:3√64 = 3√(1728/64) = 3√27 = 3

~Học tốt~

Áp dụng bđt côsi cho 3 số ta được \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}\frac{b}{c}\frac{c}{a}}=3\)

\(\Rightarrowđpcm\)

\(4,\sqrt{x}+2=x+2,\)

\(\Rightarrow\sqrt{x}+2-x-2=0\)

\(\Rightarrow x-\sqrt{x}=0\)

\(\Rightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=1\end{cases}}}\)

\(\Rightarrow x\in\left\{0;1\right\}\)

\(3=a+b+c\ge3\sqrt[3]{abc}\)\(\Leftrightarrow\)\(abc\le1\)

\(VT=\frac{a^3\left(a+1\right)+b^3\left(b+1\right)+c^3\left(c+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=\frac{a^4+b^4+c^4+a^3+b^3+c^3}{a+b+c+ab+bc+ca+abc+1}\)

\(\ge\frac{\frac{\left(a^2+b^2+c^2\right)^2}{3}+\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}}{\frac{\left(a+b+c\right)^2}{3}+5}=\frac{\frac{\frac{\left(a+b+c\right)^4}{9}}{3}+\frac{\frac{\left(a+b+c\right)^4}{9}}{3}}{8}\)

\(=\frac{\frac{\frac{3^4}{9}}{3}}{4}=\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)

đề viết gì thế bạn ?

Ta có \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\ge\frac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\left(1\right)\)

Áp dụng bất đẳng thức buniacoxki ta có :

\(\left(\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\right)\left(ab+bc+ac\right)\ge\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)^2\)

Kết hợp với (1)

=> \(\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c

Nghe mùi holder ?