\(\frac{x+y}{xyz}=\frac{1}{yz}+\frac{1}{xz}\ge\frac{4}{\left(x+y\right)z}\ge\frac{4}{\frac{\left(x+y+z\right)^2}{4}}=\frac{4}{\frac{4^2}{4}}=1\)
\(\Rightarrow x+y\ge xyz\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=1;z=2\)
cho 3 số dương x,y,z thoả mãn đk: x+y+z=4
CMR: x+y\(\ge\)xyz
cho x,y,z > 0 thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Cmr: \(\sqrt{x+yz}+\sqrt{y+xz}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)
Cho \(x\ge y\ge z>0\)
CMR : \(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2\)
Cho x,y, z ≥ 0 thỏa mãn x=y +z=1
CMR: 4(1-x)(1-y)(1-z) ≤ x+2y+z
Cho x,y,z>0 và xyz=1. cmr: x^2/(1+y) + y^2/(1+z) + z^2/(1+x) >= 3/2.?
1. Cho \(x,y,z\in\left(0,1\right)\) và \(xyz=\left(1-x\right)\left(1-y\right)\left(1-z\right)\). Cmr: \(x^2+y^2+z^2\ge\frac{3}{4}\)
2. \(\left\{{}\begin{matrix}x,y,z\ge0\\x^2+y^2+z^2+xyz=4\end{matrix}\right.\) Cmr: \(x+y+z\le3\)
3. \(x\ne-2y\). Min : \(P=\frac{\left(2x^2+13y^2-xy\right)^2-6xy+9}{\left(x+2y\right)^2}\)
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
do x,y,z≥0 nên x2≥0 , y+z≥0
áp dụng bất đẳng thức cosi cho 2 số dương \(\dfrac{x^2}{y+z}\) và y+z/4
x^2/y+z +(y+z)/4≥2\(\sqrt{\dfrac{x^2}{y+z}.\dfrac{\left(y+z\right)}{4}}\) =x (1)
y^2/x+z+(x+z)/4≥2\(\sqrt{\dfrac{y^2}{x+z}.\dfrac{x+z}{4}}\) =y (2)
z^2/y+x+(y+x)/4≥2\(\sqrt{\dfrac{z^2}{y+x}.\dfrac{y+x}{4}}\) =z (3)
từ (1)(2)(3)
➜\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)+(y+z/4)+(z+x)/4+(x+y)/4 ≥ x+y+z
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) +(a+b+c)/2 ≥x+y+z
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥ (x+y+z)/2
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥1 (vì x+y+z=2)
vậy giá trị nhỏ nhất của \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) =1
1. Cho a,b,c > 0. Cmr: a) \(\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}\le1\)
b) \(\frac{ab^2}{a^2+2b^2+c^2}+\frac{bc^2}{b^2+2c^2+a^2}+\frac{ca^2}{c^2+2a^2+b^2}\le\frac{a+b+c}{4}\)
2. Cho \(x,y,z>0;x+\frac{y}{3}+\frac{z}{5}\ge3;\frac{y}{3}+\frac{z}{5}\ge2;\frac{z}{5}\ge1.MaxP=x^2+y^2+z^2\)
3. Cho \(x>0;y\ge2;2x+y+xy\ge6.MinP=x^3+y^2\)
4. Cho \(0< \alpha< \beta< \gamma\). Giả sử x,y,z > 0 TM \(z\ge\gamma;\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}+\frac{xyz}{\alpha\beta\gamma}=4;\frac{y}{\beta}+\frac{z}{\gamma}+\frac{yz}{\beta\gamma}=3.MinP=x^3+y^3+z^3\)
