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\(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)
\(=x^4y-x^4z+y^4z-y^4x+z^4\left(x-y\right)\)
\(=xy\left(x^3-y^3\right)-z\left(x^4-y^4\right)+z^4\left(x-y\right)\)
\(=xy\left(x-y\right)\left(x^2+xy+y^2\right)-z\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)+z^4\left(x-y\right)\)
\(=\left(x-y\right)\left[xy\left(x^2+xy+y^2\right)-z\left(x^3+x^2y+xy^2+y^3\right)+z^4\right]\)
\(=\left(x-y\right)\left(x^3y+x^2y^2+xy^3-x^3z-x^2yz-xy^2z-y^3z+z^4\right)\)
\(=\left(x-y\right)\left[x^3\left(y-z\right)+x^2y\left(y-z\right)+xy^2\left(y-z\right)-z\left(y^3-z^3\right)\right]\)
\(=\left(x-y\right)\left[x^3\left(y-z\right)+x^2y\left(y-z\right)+xy^2\left(y-z\right)-z\left(y-z\right)\left(y^2+yz+z^2\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left[x^3+x^2y+xy^2-z\left(y^2+yz+z^2\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x^3+x^2y+xy^2-y^2z-yz^2-z^3\right)\)
\(=\left(x-y\right)\left(y-z\right)\left[x^3-z^3+y\left(x^2-z^2\right)+y^2\left(x-z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left[\left(x-z\right)\left(x^2+xz+z^2\right)+y\left(x-z\right)\left(x+z\right)+y^2\left(x-z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\left[x^2+xz+z^2+y\left(x+z\right)+y^2\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\frac{2\left(x^2+xz+z^2+xy+yz+y^2\right)}{2}\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\frac{x^2+2xz+z^2+x^2+xy+y^2+y^2+yz+z^2}{2}\)
\(\left(x-y\right)\left(y-z\right)\left(x-z\right)\frac{\left(x+z\right)^2+\left(x+y\right)^2+\left(y+z\right)^2}{2}\)
\(Ta\)\(có\)\(x>y>z\Rightarrow\left(x-y\right);\left(y-z\right);\left(x-z\right)>0\)
\(\left(x+z\right)^2;\left(y+z\right)^2;\left(x+y\right)^2\ge0\)
\(\Rightarrow A>o\Rightarrow A\)\(luôn\)\(dương\)
BĐT Bunhiacopxky em chưa học cô ạ
Cô cong cách nào không ạ
Nguyễn Thị Nguyệt Ánh:
Vậy thì bạn có thể chứng minh $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$ thông qua BĐT Cô-si:
Áp dụng BĐT Cô-si:
$x+y+z\geq 3\sqrt[3]{xyz}$
$xy+yz+xz\geq 3\sqrt[3]{x^2y^2z^2}$
Nhân theo vế:
$(x+y+z)(xy+yz+xz)\geq 9xyz$
$\Rightarrow \frac{xy+yz+xz}{xyz}\geq \frac{9}{x+y+z}$
hay $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$
Cho x > y > z
CMR : \(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\) luôn luôn dương
\(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)
\(A=x^4\left(y-z\right)+y^4\left(z-x\right)-z^4\left[\left(y-z\right)+\left(z-x\right)\right]\)
\(A=x^4\left(y-z\right)-z^4\left(y-z\right)+y^4\left(z-x\right)-z^4\left(z-x\right)\)
\(A=\left(y-z\right)\left(x^4-z^4\right)+\left(z-x\right)\left(y^4-z^4\right)\)
\(A=\left(y-z\right)\left(x-z\right)\left(x+z\right)\left(x^2+z^2\right)-\left(x-z\right)\left(y-z\right)\left(y+z\right)\left(y^2+z^2\right)\)
\(A=\left(y-z\right)\left(x-z\right)\left(x^3+xz^2+x^2z+z^3-y^3-yz^2-y^2z-z^3\right)\)
\(A=\left(y-z\right)\left(x-z\right)\left(x-y\right)\left(x^2+xy+y^2+z^2+zx+yz\right)\)
\(A=\frac{1}{2}\left(x-y\right)\left(y-z\right)\left(x-z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\right]\)
Vì \(x>y>z\Rightarrow A>0\)
a ) Đặt A = \(\frac{-a+b+c}{2a}+\frac{a-b+c}{2b}+\frac{a+b-c}{2c}=\frac{1}{2}\left(-1+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}-1+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}-1\right)\)
\(=\frac{1}{2}\left(\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}-3\right)\)
Do a ; b ; c > 0 , áp dụng BĐT Cô - si cho các cặp số dương , ta có :
\(A\ge\frac{1}{2}\left[2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{b}{c}.\frac{c}{b}}+2\sqrt{\frac{a}{c}.\frac{c}{a}}-3\right]=\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
b ) \(P=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{x^2}{xy+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{xz+yz}\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\frac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\frac{3}{2}\)
( áp dụng BĐT Cauchy - Schwarz )
Dấu " = " xảy ra \(\Leftrightarrow x=y=z\)
Theo Cauche có:
\(\left(x+x+y+z\right)\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge4\sqrt[4]{x^2yz}.4\sqrt[4]{\frac{1}{x^2.y.z}}=16\)
=> \(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{16}{2x+y+z}\). Tương tự có:
\(\frac{2}{y}+\frac{1}{x}+\frac{1}{z}\ge\frac{16}{x+2y+z}\) và \(\frac{2}{z}+\frac{1}{y}+\frac{1}{x}\ge\frac{16}{x+y+2z}\)
=> \(16.\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\le\frac{2}{x}+\frac{1}{y}+\frac{1}{z}+\frac{2}{y}+\frac{1}{x}+\frac{1}{z}+\frac{2}{z}+\frac{1}{x}+\frac{1}{y}\)
\(16.\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\le4.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=4.4=16\)
Chia cả 2 vế cho 16 => ĐPCM