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P = x(x/2+1/yz) + y(y/2+1/zx) + z(z/2+1/xy)
= ½ [x(xyz +2)/(yz) + y(xyz +2)/(xz) + z(xyz +2)/(xy)]
= ½ (xyz +2)[x/(yz) + y/(xz) + z/(xy)] ≥ ½ (xyz +2).3 /³√(xyz)
Lại có: xyz + 2 = xyz + 1 +1 ≥ 3 ³√(xyz)
Suy ra:
P = ½ (xyz +2)[x/(yz) + y/(xz) + z/(xy)] ≥ ½ (xyz +2).3 /³√(xyz)
≥ 3/2 .3 ³√(xyz)/ ³√(xyz) = 9/2
Vậy P min = 9/2
Dấu = xra khi x = y = z = 1
Bài 1:
Ta có
A =x/(x+1) +y/(y+1)+z/(z+1)
A= 1- 1/(x+1)+1-1/(y+1) +1-1/(z+1)
A=3- [1/(x+1)+1/(y+1) +1/(z+1) ]
B = 1/(x+1)+1/(y+1) +1/(z+1)
Đặt x+1=a; y+1=b;z+1 =c
=>a+b+c=4
4B=4(1/a+1/b+1/c)
B= (a+b+c) (1/a+1/b+1/c)
4B =3+(a/b+b/a) +(a/c+c/a)+(b/c+c/a)
Từ (a-b)^2 ≥ 0 =>a^2+b^2 ≥ 2ab chia 2 vế cho ab
=> a/b+b/a ≥2 dấu "=" khi a=b
Tương tự có
a/c+c/a ≥2 ;b/c+c/b ≥2
=>4B ≥3+2+2+2=9
=>B ≥ 9/4
=>A ≤ 3-9/4 = 3/4
Vậy max A =3/4 khi a=b=c
=>x=y=z =1/3
Bài 2:
Giúp tui nha
\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)
\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)
\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
\(P+3=x+\left(y^2+1\right)+\left(z^3+1+1\right)\ge x+2y+3z\)
\(\Rightarrow P\ge x+2y+3z-3\)
\(6=\dfrac{1}{x}+\dfrac{4}{2y}+\dfrac{9}{3z}\ge\dfrac{\left(1+2+3\right)^2}{x+2y+3z}\)
\(\Rightarrow x+2y+3z\ge6\Rightarrow P\ge3\)
Dấu "=" xảy ra khi \(x=y=z=1\)
