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Đặt x+y−z=a;x−y+z=b;−x+y+z=cx+y−z=a;x−y+z=b;−x+y+z=c thì a + b + c = x + y + z
A=(a+b+c)3−a3−b3−c3A=(a+b+c)3−a3−b3−c3
=(a+b+c−a)[(a+b+c)2+a(a+b+c)+a2]−(b3+c3)=(a+b+c−a)[(a+b+c)2+a(a+b+c)+a2]−(b3+c3)
=(b+c)[a2+b2+c2+2(ab+bc+ca)+(a2+ab+ac)+a2]−(b+c)(b2−bc+c2)=(b+c)[a2+b2+c2+2(ab+bc+ca)+(a2+ab+ac)+a2]−(b+c)(b2−bc+c2)=(b+c)[3a2+b2+c2+3ab+2bc+3ac−b2+bc−c2]=(b+c)[3a2+b2+c2+3ab+2bc+3ac−b2+bc−c2]
=(b+c)(3a2+3ab+3bc+3ca)=(b+c)(3a2+3ab+3bc+3ca)
=(b+c)(3a(a+b)+3c(a+b))=3(a+b)(b+c)(c+a)
Bài 3:
Áp dụng BĐT Cauchy cho các số dương ta có:
\(\frac{1}{x}+\frac{x}{4}\geq 2\sqrt{\frac{1}{4}}=1\)
\(\frac{1}{y}+\frac{y}{4}\geq 2\sqrt{\frac{1}{4}}=1\)
\(\frac{1}{z}+\frac{z}{4}\geq 2\sqrt{\frac{1}{4}}=1\)
Cộng theo vế các BĐT vừa thu được ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{x+y+z}{4}\geq 3\)
\(\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 3-\frac{x+y+z}{4}\geq 3-\frac{6}{4}\) (do \(x+y+z\leq 6\) )
\(\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z=2\)
Bài 4:
Áp dụng BĐT Cauchy cho 3 số dương:
\(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq 3\sqrt[3]{\frac{x}{y}.\frac{y}{z}.\frac{z}{x}}=3\sqrt[3]{1}=3\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z\)
a ) \(\frac{1}{\left(x-y\right)\left(y-z\right)}+\frac{1}{\left(y-z\right)\left(z-x\right)}+\frac{1}{\left(z-x\right)\left(x-y\right)}\)
= \(\frac{z-x}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{x-y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
= \(\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
b ) \(\frac{4}{\left(y-x\right)\left(z-x\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)
= \(\frac{-4}{\left(y-x\right)\left(x-z\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)
= \(\frac{-4\left(y-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(x-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)
= \(\frac{-4y+4z+3x-3z+3y-3x}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{z-y}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)
= \(\frac{-\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{-1}{\left(x-z\right)\left(y-z\right)}=\frac{1}{\left(x-z\right)\left(x-y\right)}\)
Chúc bạn học tốt !!!
\(x^3+y^3+z^3=3xyz\\ \Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\\ \Rightarrow x=y=z\)
tháy vào M ta có:
\(M=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\\ M=\left(1+1\right)\left(1+1\right)\left(1+1\right)=\left(1+1\right)^3=8\)
\(x^3+y^3+z^3=3xyz\Leftrightarrow x+y+z=0\)(chỗ này nếu k hiểu thì hỏi mk giải thích cho)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\z+x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=-x-y\\x=-y-z\\y=-z-x\end{matrix}\right.\)
Thay vào M ta có:
\(M=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)
\(=\left(1+\dfrac{-y-z}{y}\right)\left(1+\dfrac{-z-x}{z}\right)\left(1+\dfrac{-x-y}{x}\right)\)
\(=\left(1-1-\dfrac{z}{y}\right)\left(1-1-\dfrac{x}{z}\right)\left(1-1-\dfrac{y}{x}\right)\)
\(=\left(-\dfrac{z}{y}\right)\left(-\dfrac{x}{z}\right)\left(-\dfrac{y}{x}\right)\)
\(=-1\)