Cho \(x>y>z\) CMR \(x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(z-y\right)>0\)
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Những câu hỏi liên quan
16 tháng 8 2019
Áp dụng bất đẳng thức Cauchy :
\(\frac{x^4}{y^2\left(x+z\right)}+\frac{y^2}{2x}+\frac{x+z}{4}\ge3\sqrt[3]{\frac{x^4\cdot y^2\cdot\left(x+z\right)}{y^2\cdot\left(x+z\right)\cdot2x\cdot4}}=3\sqrt[3]{\frac{x^3}{8}}=\frac{3x}{2}\)
Tương tự ta cũng có :
\(\frac{y^4}{z^2\left(x+y\right)}+\frac{z^2}{2y}+\frac{x+y}{4}\ge\frac{3y}{2}\)
\(\frac{z^4}{x^2\left(y+z\right)}+\frac{x^2}{2z}+\frac{y+z}{4}\ge\frac{3z}{2}\)
Cộng theo vế ta được :
\(VT+\left(\frac{y^2}{2x}+\frac{z^2}{2y}+\frac{x^2}{2z}\right)+\frac{2\left(x+y+z\right)}{4}\ge\frac{3x}{2}+\frac{3y}{2}+\frac{3z}{2}\)
\(\Leftrightarrow VT+\frac{1}{2}\left(\frac{y^2}{x}+\frac{z^2}{y}+\frac{x^2}{z}\right)+\frac{1}{2}\left(x+y+z\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Leftrightarrow VT+\frac{1}{2}\cdot\frac{\left(x+y+z\right)^2}{x+y+z}+\frac{1}{2}\left(x+y+z\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Leftrightarrow VT+\frac{1}{2}\left(x+y+z\right)+\frac{1}{2}\left(x+y+z\right)\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Leftrightarrow VT\ge\frac{x+y+z}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 9 2021
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
\(\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)^3}\)
\(\Rightarrow3\left(xy+yz+zx\right)^3\le\left(\dfrac{9}{8}\right)^2\)
\(\Rightarrow\left(xy+yz+zx\right)^3\le\dfrac{27}{64}\)
\(\Rightarrow xy+yz+zx\le\dfrac{3}{4}\)
AH
Akai Haruma
Giáo viên
21 tháng 5 2021
Hầy mình không nghĩ lớp 7 đã phải làm những bài biến đổi như thế này. Cái này phù hợp với lớp 8-9 hơn.
1.
Đặt $x^2-y^2=a; y^2-z^2=b; z^2-x^2=c$.
Khi đó: $a+b+c=0\Rightarrow a+b=-c$
$\text{VT}=a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3$
$=(-c)^3-3ab(-c)+c^3=3abc$
$=3(x^2-y^2)(y^2-z^2)(z^2-x^2)$
$=3(x-y)(x+y)(y-z)(y+z)(z-x)(z+x)$
$=3(x-y)(y-z)(z-x)(x+y)(y+z)(x+z)$
$=3.4(x-y)(y-z)(z-x)=12(x-y)(y-z)(z-x)$
Ta có đpcm.
AH
Akai Haruma
Giáo viên
21 tháng 5 2021
Bài 2:
Áp dụng kết quả của bài 1:
Mẫu:
$(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3=3(x-y)(y-z)(z-x)(x+y)(y+z)(z+x)=3(x-y)(y-z)(z-x)(1)$
Tử:
Đặt $x-y=a; y-z=b; z-x=c$ thì $a+b+c=0$
$(x-y)^3+(y-z)^3+(z-x)^3=a^3+b^3+c^3$
$=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3=3abc$
$=3(x-y)(y-z)(z-x)(2)$
Từ $(1);(2)$ suy ra \(\frac{(x-y)^3+(y-z)^3+(z-x)^3}{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}=1\)
8 tháng 2 2019
\(4\left(y-z\right)\left(z-x\right)+4\left(z-y\right)\left(x-y\right)+4\left(x-z\right)\left(y-z\right)=0\)
\(\Leftrightarrow4yz-4xy-4xz+4x^2+4xz-4yz-4xy+4y^2+4xy-4xz-4yz+4z^2=0\)\(\Leftrightarrow4x^2-4xy-4xz+4y^2-4yz+4z^2=0\)
\(\Leftrightarrow4\left(x^2-xy-xz+y^2-yz+z^2\right)=0\)
\(\Leftrightarrow x^2+y^2+z^2-xy-xz-yz=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)
\(\Leftrightarrow x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-z\right)^2=0\\\left(y-z\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-z=0\\y-z=0\end{matrix}\right.\Leftrightarrow x=y=z\)
3 tháng 12 2016
\(x+y+z+\sqrt{xyz}=4\)
\(\Leftrightarrow xyz=\left(4-x-y-z\right)^2\)
\(\Leftrightarrow xyz=16+x^2+y^2+z^2-8x-8y-8z+2xy+2xz+yz\)
\(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left(16-4y-4z+yz\right)}=\sqrt{16x-4xy-4xz+xyz}\)
\(=\sqrt{16x-4xy-4xz+16+x^2+y^2+z^2-8x-8y-8z+2xy+2yz+2xz}\)
\(=\sqrt{8x-2xy-2xz+2yz+x^2+y^2+z^2-8y-8z+16}\)
\(=\sqrt{\left(-x+y+z-4\right)^2}=\left|y+z-x-4\right|=\left|y+z-x-\left(x+y+z+\sqrt{xyz}\right)\right|\)
\(=\left|-2x-\sqrt{xyz}\right|=2x+\sqrt{xyz}\) (Vì x > 0)
Tương tự : \(\sqrt{y\left(4-z\right)\left(4-x\right)}=2y+\sqrt{xyz}\) , \(\sqrt{z\left(4-x\right)\left(4-y\right)}=2z+\sqrt{xyz}\)
Suy ra \(B=2x+2y+2z+2\sqrt{xyz}=2\left(x+y+z+\sqrt{xyz}\right)=2.4=8\)
Sửa đề z^4(z-y) thành z^4(x-y)
Đặt \(A=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)
\(=x^4\left(y-x+x-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)
\(=-x^4\left(x-y\right)+x^4\left(x-z\right)-y^4\left(x-z\right)+z^4\left(x-y\right)\)
\(=\left(x-y\right)\left(z^4-x^4\right)+\left(x-z\right)\left(x^4-y^4\right)\)
\(=\left(x-y\right)\left(z^2+x^2\right)\left(z^2-x^2\right)+\left(x-z\right)\left(x^2+y^2\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(z^2+x^2\right)\left(x+z\right)\left(z-x\right)+\left(x-z\right)\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(z-x\right)\left[\left(z^2+x^2\right)\left(x+z\right)-\left(x^2+y^2\right)\left(x+y\right)\right]\)
\(=\left(x-y\right)\left(z-x\right)\left(xz^2+z^3+x^3+x^2z-x^3-x^2y-xy^2-y^3\right)\)
\(=\left(x-y\right)\left(z-x\right)\left[x^2\left(z-y\right)+x\left(z^2-y^2\right)+\left(z^3-y^3\right)\right]\)
\(=\left(x-y\right)\left(z-x\right)\left(z-y\right)\left[x^2+x\left(z+y\right)+\left(z^2+yz+y^2\right)\right]\)
\(=\left(x-y\right)\left(x-z\right)\left(y-z\right)\left(x^2+xz+xy+z^2+yz+y^2\right)\)
\(=\frac{1}{2}\left(x-y\right)\left(x-z\right)\left(y-z\right)\left(2x^2+2y^2+2z^2+2xy+2yz+2xz\right)\)
\(=\frac{1}{2}\left(x-y\right)\left(x-z\right)\left(y-z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\right]\)
Vì \(x>y>z\Rightarrow\hept{\begin{cases}x-y>0\\x-z>0\\y-z>0\end{cases}}\) và \(\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\ge0\)
=>....