Cho x, y, z > 0 và xyz = 1
Tìm GTLN của \(P=\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\)
Cho các số thực dương x,y,z thỏa mãn xyz ≥ 1.Tìm GTNN của \(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)
\(x,y,z>0\)
Áp dụng BĐT Caushy cho 3 số ta có:
\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)
\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)
\(=\dfrac{\left(x^3-1\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)}+\dfrac{\left(y^3-1\right)^2}{\left(x+y^2+z\right)\left(y^3-1\right)}+\dfrac{\left(z^3-1\right)^2}{\left(x+y+z^2\right)\left(x^3-1\right)}\)
Áp dụng BĐT Caushy-Schwarz ta có:
\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)
\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)
\(P=0\Leftrightarrow x=y=z=1\)
Vậy \(P_{min}=0\)
Cho x,y,z>0 /xyz=8.
Tìm min P= \(\dfrac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\dfrac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\dfrac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
Cho x, y, z > 0 và xyz=1. CMR :
\(\dfrac{x^2}{1+y}+\dfrac{y^2}{1+z}+\dfrac{z^2}{1+z}\ge\dfrac{3}{2}\)
Đề sai nhé, \(\dfrac{z^2}{x+1}\) mới đúng nha
\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3}\left(\text{Svácxơ}\right)\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Ta có: \(x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow x+y+z+3\le2\left(x+y+z\right)\)
a)Chứng minh x3 + y3 ≥xy(x+y) với x,y≥0
b)Cho x,y,z>0 thỏa mãn xyz=1
CMR:\(\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\le1\)
Lời giải:
a. Xét hiệu:
$x^3+y^3-xy(x+y)=(x^3-x^2y)-(xy^2-y^3)=x^2(x-y)-y^2(x-y)$
$=(x-y)(x^2-y^2)=(x-y)^2(x+y)\geq 0$ với mọi $x,y\geq 0$
$\Rightarrow x^3+y^3\geq xy(x+y)$
Dấu "=" xảy ra khi $x=y$
b.
Áp dụng BĐT phần a vô:
$x^3+y^3\geq xy(x+y)$
$\Rightarrow x^3+y^3+1\geq xy(x+y)+1=xy(x+y)+xyz=xy(x+y+z)$
$\Rightarrow \frac{1}{x^3+y^3+1}\leq \frac{1}{xy(x+y+z)}=\frac{xyz}{xy(x+y+z)}=\frac{z}{x+y+z}$
Hoàn toàn tương tự với các phân thức còn lại suy ra:
$\text{VT}\geq \frac{z}{x+y+z}+\frac{x}{x+y+z}+\frac{y}{x+y+z}=1$
Ta có đpcm
Dấu "=" xảy ra khi $x=y=z=1$
a, Cho x, y, z > 0 \(\in[0,1]\). Chứng minh:
\(\dfrac{x}{yz+1}+\dfrac{y}{xz+1}+\dfrac{z}{xy+1}< 2\)
b, x, y, z > 0 : xyz = 1. Chứng minh:
\(\dfrac{1}{x^2+2y+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\le2\)
với x,y,z>0 và \(x+y+z\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
chứng minh đẳng thức \(x+y+z\ge\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\)
\(\Rightarrow\left(x+y+z\right)^2\ge\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\ge3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{3\left(x+y+z\right)}{xyz}\Rightarrow x+y+z\ge\dfrac{3}{xyz}\)
\(x+y+z=\dfrac{x+y+z}{3}+\dfrac{2\left(x+y+z\right)}{3}\ge\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{2}{3}.\dfrac{3}{xyz}\ge\dfrac{1}{3}\left(\dfrac{9}{x+y+z}\right)+\dfrac{2}{xyz}=\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\left(đpcm\right)\)
\(dấu"="xảy\) \(ra\Leftrightarrow x=y=z=1\)
Cho x, y, z >0 thỏa mãn : xyz=1. CMR :
\(\dfrac{\sqrt{1+x^3+y^3}}{xy}+\dfrac{\sqrt{1+y^3+z^3}}{yz}+\dfrac{\sqrt{1+z^2+x^2}}{xz}\ge3\sqrt{3}\)
\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)
\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)
\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)
=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)
Cho các số dương x,y,z thỏa mãn xyz=1. Tìm Min \(P=\dfrac{\sqrt{1+x^3+y^3}}{xy}+\dfrac{\sqrt{1+y^3+z^3}}{yz}+\dfrac{\sqrt{1+z^3+x^3}}{zx}\)
\(P\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}+\dfrac{\sqrt{3\sqrt[3]{y^3z^3}}}{yz}+\dfrac{\sqrt{3\sqrt[3]{z^3x^3}}}{zx}\)
\(P\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.zx}}}=3\sqrt{3}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Ta có bất đẳng thức sau \(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0.\)
Do đó:
\(P=\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}\)
\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\dfrac{1}{\sqrt{xy}}\cdot\dfrac{1}{\sqrt{yz}}\cdot\dfrac{1}{\sqrt{zx}}}=3\sqrt{3}\)
Đẳng thức xảy ra khi $x=y=z=1.$
Áp dụng BĐT AM-GM:
\(P=\dfrac{\sqrt{1+x^3+y^3}}{xy}+\dfrac{\sqrt{1+y^3+z^3}}{yz}+\dfrac{\sqrt{1+z^3+x^3}}{zx}\)
\(\ge\dfrac{\sqrt{3xy}}{xy}+\dfrac{\sqrt{3yz}}{yz}+\dfrac{\sqrt{3zx}}{zx}\)
\(=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)
\(\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)
\(minP=3\sqrt{3}\Leftrightarrow x=y=z\)
Cho x,y,z là các số thực dương. Chứng minh rằng:
\(\dfrac{1}{x^3+y^3+xyz}+\dfrac{1}{y^3+z^3+xyz}+\dfrac{1}{z^3+x^3+xyz}\le\dfrac{1}{xyz}\)
do x,y,z là các số dương nên
\(x^2-xy+y^2\ge xy\Leftrightarrow x^3+y^3\ge xy\left(x+y\right)\)
tương tự ta cũng có : \(y^3+z^3\ge yz\left(y+z\right)\)
\(z^3+x^3\ge zx\left(z+x\right)\)
\(\Rightarrow\Sigma\dfrac{1}{x^3+y^3+xyz}\le\Sigma\dfrac{1}{xy\left(x+y+z\right)}=\dfrac{1}{x+y+z}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)\)
\(=\dfrac{1}{x+y+z}\left(\dfrac{x+y+z}{xyz}\right)=\dfrac{1}{xyz}\left(đpcm\right)\)