Ta có: 

\(x^2+y^2=\left(x+y\right)^2-2xy=a^2-2b\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=a^3-3ab\)

\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left(a^2-2b\right)^2-2b^2\)

\(=a^4-4a^2b+4b^2-2b^2=a^4-4a^2b+2b^2\)

\(x^5+y^5=\left(x+y\right)^5-\left(5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)\)

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

\(=a^5-5\left(a^3-3ab\right)b-10ab^2\)

\(=a^5-5a^3b+15ab^2-10ab^2\)

\(=a^5-5a^3b+5ab^2\)

\(x^2+y^2=\left(x+y\right)^2-2xy=a^2-2b\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=a^3-3ab\)

\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left[\left(x+y\right)^2-2xy\right]^2-2x^2y^2=\left(a^2-2b\right)^2-2b^2\)

\(=a^2-4a^2b+2b^2\)

\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)=\left(a^2-2b\right)\left(a^3-3ab\right)-ab^2\)

Có: \(x^2-x+\frac{23}{4}=x^2-2.\frac{1}{2}.x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+\frac{23}{4}\)

     \(=\left(x-\frac{1}{2}\right)^2+\frac{11}{2}\ge\frac{11}{2}\)

Vậy Min = 11/2 khi x - 1/2 = 0 => x = 1/2

\(P=\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)=\frac{\left(a+1\right)\left(b+1\right)}{ab}\)

Áp dụng Cosi 3 số

\(a+1=a+a+b\ge3\sqrt[3]{a^2b}\)

\(a+1=b+b+a\ge3\sqrt[3]{ab^2}\)

Nhận lại 3 BĐT trên theo vế:

\(\left(a+1\right)\left(b+1\right)\ge9ab\)

\(\Leftrightarrow\frac{\left(a+1\right)\left(b+1\right)}{ab}\ge9\)

\(\Leftrightarrow P\ge9\)

Đẳng thức xảy ra khi a=b=c

dùng BĐT Côsy

áp dụng cô si ta có 
a³/b + ab ≥ 2a² 
b³/c + bc ≥ 2b² 
c³/a + ac ≥ 2c² 
+ + + 3 cái lại 
=> a³/b + b³/c + c³/a ≥ 2a² + 2b² + 2c² - ab - ac - bc 
mặt khác ta có 
ab + bc + ac ≤ a² + b² + c² (cái này chứng minh dễ dàng nhé) 

Cảm ơn bạn nhé

\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{x+z}+1\right)+\left(\frac{z}{x+y}+1\right)-3\)

\(=\frac{x+y+z}{y+z}+\frac{x+y+z}{x+z}+\frac{x+y+z}{x+y}-3=\left(x+y+z\right).\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)-3\)

\(=\frac{1}{2}\left[\left(x+y\right)+\left(y+z\right)+\left(x+z\right)\right]\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)-3\ge\frac{9}{2}-3=\frac{3}{2}\left(đpcm\right)\)

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

Áp dụng Cô-Si cho các số không âm:

\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2;\frac{y}{z}+\frac{z}{y}\ge2\sqrt{\frac{y}{z}.\frac{z}{y}}=2;\frac{x}{z}+\frac{z}{x}\ge2\sqrt{\frac{x}{z}.\frac{z}{x}}=2\)

Cộng theo vế các bất đẳng thức ta được: \(\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\ge2+2+2=6\)

Xem lại đề...............

ban chuyen ve tao hang dang thuc thu 2 . sau do dung co si hoac bunhia ngc .( neu dung cosi thi them tri tuyet doi , con d amung bunhia thi ko lo duong hay am

Xét hiệu: \(\frac{a^2}{b^2}+\frac{b^2}{a^2}+4-3\left(\frac{a}{b}+\frac{b}{a}\right)=\left(\frac{a}{b}+\frac{b}{a}\right)^2+2-3\left(\frac{a}{b}+\frac{b}{a}\right)\left(1\right)\) 

Đặt \(\frac{a}{b}+\frac{b}{a}=A\) , (1) trở thành: \(A^2-3A+2=A^2-A-2A+2=A\left(A-1\right)-2\left(A-1\right)=\left(A-1\right)\left(A-2\right)\)

+Nếu a,b cùng dấu ,ta có:  \(A=\frac{a}{b}+\frac{b}{a}\) \(\ge2\) (c/m = biến đổi tương đương)

Do đó \(\left(A-1\right)\left(A-2\right)\ge0\),Dấu "=" xảy ra <=> a=b

+Nếu a,b trái dấu ,ta có: \(A=\frac{a}{b}+\frac{b}{a}\le-2\)

do đó \(\left(A-1\right)\left(A-2\right)\ge0\),Dấu "=" xảy ra <=> a=-b

Từ đó suy ra đpcm