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Áp dụng BĐT AM-GM: \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)
a) Áp dụng: \(VT\ge\frac{\left(a+b+c\right)^2}{3}.\frac{9}{2\left(a+b+c\right)}=\frac{3}{2}\left(a+b+c\right)\)
b) \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{x+y+z+3}=\frac{3}{4}\)
Lời giải:
Vì $xy+yz+xz=1$ nên:
\(x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\)
\(y^2+1=y^2+xy+yz+xz=(y+x)(y+z)\)
\(z^2+1=z^2+xy+yz+xz=(z+y)(z+x)\)
Do đó:
\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{1+z^2}=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)}\)
\(=\frac{x(y+z)+y(x+z)+z(x+y)}{(x+y)(y+z)(x+z)}=\frac{2(xy+yz+xz)}{(x+y)(y+z)(x+z)}=\frac{2}{\sqrt{(x+y)^2(y+z)^2(x+z)^2}}\)
\(=\frac{2}{\sqrt{(x+y)(x+z)(y+z)(y+x)(z+x)(z+y)}}=\frac{2}{\sqrt{(x^2+1)(y^2+1)(z^2+1)}}\) (đpcm)
Lời giải:
Vì $xy+yz+xz=1$ nên:
\(x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\)
\(y^2+1=y^2+xy+yz+xz=(y+x)(y+z)\)
\(z^2+1=z^2+xy+yz+xz=(z+y)(z+x)\)
Do đó:
\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{1+z^2}=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)}\)
\(=\frac{x(y+z)+y(x+z)+z(x+y)}{(x+y)(y+z)(x+z)}=\frac{2(xy+yz+xz)}{(x+y)(y+z)(x+z)}=\frac{2}{\sqrt{(x+y)^2(y+z)^2(x+z)^2}}\)
\(=\frac{2}{\sqrt{(x+y)(x+z)(y+z)(y+x)(z+x)(z+y)}}=\frac{2}{\sqrt{(x^2+1)(y^2+1)(z^2+1)}}\) (đpcm)
\(VT=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(=2+\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\)
Bài toán trở thành \(\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\ge\frac{x+y+z}{3\sqrt{xyz}}\)
Áp dụng bất đẳng thức AM-GM:
\(\frac{z}{x}+\frac{z}{y}+\frac{z}{z}\ge3\sqrt[3]{\frac{z^3}{xyz}}=\frac{3z}{\sqrt[3]{xyz}}\)
Tương tự:
\(\frac{y}{x}+\frac{y}{z}+\frac{y}{y}\ge\frac{3y}{\sqrt[3]{xyz}}\)
\(\frac{x}{z}+\frac{x}{y}+\frac{x}{x}\ge\frac{3x}{\sqrt[3]{xyz}}\)
\(\Leftrightarrow VT+3\ge3+\frac{3}{\sqrt[3]{xyz}}\left(x+y+z\right)\)
\(\Leftrightarrow VT\ge\frac{3\left(x+y+z\right)}{\sqrt[3]{xyz}}\)\(\ge\frac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Is it true?
Lời giải:
Tập A sửa lại thành \(A=\left\{\frac{1}{6};\frac{1}{12};\frac{1}{20}; \frac{1}{30};....;\frac{1}{420}\right\}\)
Ta thấy:
\(\frac{1}{6}=\frac{1}{2.3}\)
\(\frac{1}{12}=\frac{1}{3.4}\)
\(\frac{1}{20}=\frac{1}{4.5}\)
.....
\(\frac{1}{420}=\frac{1}{20.21}\)
Do đó công thức tổng quát của các phần tử thuộc tập A là \(\frac{1}{x(x+1)}|x\in \mathbb{N}; 2\leq x\leq 20\)
Đáp án D.
ĐKXĐ : \(x\ne0;1;-1\)
Ta có :
\(C=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x+1}{x^2-1}\right).\frac{x+2006}{x}\)
\(=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-4x+1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+2006}{x}\)
\(=\frac{\left(x+1-x+1\right)\left(x+1+x-1\right)+x^2-4x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2006}{x}\)
\(=\frac{2x+x^2-4x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2006}{x}\)
\(=\frac{\left(x-1\right)^2\left(x+2006\right)}{\left(x-1\right)\left(x+1\right).x}\)
\(=\frac{\left(x-1\right)\left(x+2006\right)}{x\left(x-1\right)}\)
Vậy....