Ta cần chứng minh \((1+a)(1+b)(1+c) \geq (1+\sqrt[3]{abc})^3\)

\(\Leftrightarrow 1+abc+ab+bc+ca+a+b+c \geq 1+3\sqrt[3]{(abc)^2}+3\sqrt[3]{abc}+abc\)

\(\Leftrightarrow ab+bc+ca+a+b+c \geq 3\sqrt[3]{(abc)^2}+3\sqrt[3]{abc}\)

Đúng theo BĐT AM-GM. Áp dụng vào ta có:

\(\left(1+\frac{1}{x} \right)\left(1+\frac{1}{y} \right)\left(1+\frac{1}{z} \right)=\dfrac{(1+x)(1+y)(1+z)}{xyz} \geq \dfrac{(1+\sqrt[3]{xyz})^3}{xyz} \geq 64\)
Từ \(x+y+z=1\Rightarrow xyz\le \frac{1}{27}\)

\(\Rightarrow \dfrac{(1+\sqrt[3]{xyz})^3}{xyz}=\bigg(\dfrac{1}{\sqrt[3]{xyz}}+1\bigg)^3 \geq 64\)

Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{3}\)

Áp dụng trực tiếp BĐT AM-GM ta có:

\(1+\dfrac{1}{x}=\dfrac{1}{x}\left(x+y+z+x\right)\ge\dfrac{1}{x}4\sqrt[4]{x^2yz}\)

\(\Rightarrow1+\dfrac{1}{x}\ge\dfrac{4}{x}\sqrt[4]{\dfrac{x^4yz}{x^2}}=4\sqrt[4]{\dfrac{yz}{x^2}}\)

Tương tự ta có: \(1+\dfrac{1}{y}\ge4\sqrt[4]{\dfrac{xz}{y^2}};1+\dfrac{1}{z}\ge4\sqrt[4]{\dfrac{xy}{z^2}}\)

\(\Rightarrow\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\ge4\sqrt[4]{\dfrac{yz}{x^2}}4\sqrt[4]{\dfrac{xz}{y^2}}4\sqrt[4]{\dfrac{xy}{z^2}}=64\)

Còn tỉ tỉ cách nữa đây, cần thì nhắn tin ==

d)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)

=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)

=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)

a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

=0

c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{1}{xyz}\)

b: \(M=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=\dfrac{a+b+c}{abc}=0\)

c: \(B=\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(x-z\right)\left(y-z\right)}-\dfrac{x}{\left(x-z\right)\left(x-y\right)}\)

\(=\dfrac{y\left(x-z\right)-z\left(x-y\right)-x\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{xy-yz-xz+zy-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)

a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)

b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)

Ta có

\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=1+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xyz}\)

áp dụng bất đẳng thức CS ta có

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=9\) ;

\(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\ge\dfrac{9}{xy+yz+xz}\)

ta có đánh giá : \(xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{1}{3}\)

\(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{27}\Rightarrow\dfrac{1}{xyz}\ge27\)

\(\Rightarrow1+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xyz}\ge1+9+27+27=64\)

\(\Rightarrowđpcm\)

Ta có: \(\left(x+z\right)\left(y+z\right)=1\)

\(\Rightarrow\left(x+z\right)^2\left(y+z\right)^2=1\)

\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(y+z\right)^2}+\dfrac{1}{\left(z+x\right)^2}=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x+z\right)^2\left(y+z\right)^2}{\left(y+z\right)^2}+\dfrac{\left(x+z\right)^2\left(y+z\right)^2}{\left(z+x\right)^2}\)

\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z\right)^2+\left(y+z\right)^2\)

\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z\right)^2-2\left(x+z\right)\left(y+z\right)+\left(y+z\right)^2+2\) (Vì: (x+z)(y+z)=1 =>2(x+z)(y+z)=2 )

\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x+z-y-z\right)^2+2\)

\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2+2\)

Áp dụng bất đẳng thức Cauchy, ta có :

\(\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2\ge2\sqrt{\dfrac{1}{\left(x-y\right)^2}\cdot\left(x-y\right)^2}=2\cdot1=2\)

\(\Rightarrow P=\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2+2\ge2+2=4\)

Vậy \(MinP=4\) khi \(x-y=1\); \(y+z=\dfrac{\sqrt{5}-1}{2}\); \(x+z=\dfrac{2}{\sqrt{5}-1}\)

Ta có: \(x-y-z=0\)

\(\Rightarrow x-y=z\)

\(x-z=y\)

\(y+z=x\)

\(\Rightarrow B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)

\(=\dfrac{x-z}{x}.\dfrac{-\left(y-x\right)}{y}.\dfrac{z+y}{z}\)

\(=\dfrac{y}{x}.-\dfrac{z}{y}.\dfrac{z}{x}=-1\)

\(\Rightarrow B=-1\)