Chứng minh các bất đẳng thức sau:
a) \(\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)
b) \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)
c) \(a^2+2b^2+c^2\ge2ab-2bc\)
d) \(x^2+y^2+z^2+\dfrac{3}{4}\ge x+y+z\)
e) \(a^2+b^2\ge\left(a+b\right)^2\ge4ab\)
f) \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
1. Chứng minh các bất đẳng thức sau:
a. \(a^2+b^2+c^2\ge ab+bc+ca\)
b. \(a^2+b^2+c^2+d^2\ge ab+bc+cd+da\)
c. \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
2. Cho x,y,z không âm. Cmr: \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\)
3. Cho a+b+c=1. Cm: \(a^2+b^2+c^2\ge\dfrac{1}{3}\)
Chứng minh:
a. \(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
b. \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)
c. \(a^2+b^2+c^2+d^2\ge ab+ac+ad\)
d. \(\left(a+b+c\right)\left(x+y+z\right)\ge3\left(ax+by+cz\right)\) (Gợi ý: Bất đẳng thức Trê-bư-xếp)
Giúp em với! <3
cho a,b,c,x,y,z > 0 chứng minh rằng : \(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\ge\dfrac{\left(x+y+z\right)^2}{a+b+c}\)
Cho các số x, y, z dương thỏa mãn: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3\)
Cmr: \(\dfrac{1}{\left(2x+y+z\right)^2}+\dfrac{1}{\left(2y+z+x\right)^2}+\dfrac{1}{\left(2z+x+y\right)^2}\ge\dfrac{3}{16}\)
chứng minh rằng : \(x^3+y^3+z^3\ge\left(\dfrac{x+y}{2}\right)^3+(\dfrac{y+z}{2})^3+\left(\dfrac{z+x}{2}\right)^3\)
ta có \(A=\dfrac{1}{1+\dfrac{bc}{a}}+\dfrac{1}{1+\dfrac{ca}{b}}+\dfrac{1}{1+\dfrac{ab}{c}}\)
đặt \(\sqrt{\dfrac{bc}{a}};\sqrt{\dfrac{ca}{b}};\sqrt{\dfrac{ab}{c}}=\left(x;y;z\right)\) =>xy+yz+zx=1
ta có A=\(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{1}{1+z^2}\)
ta cần chứng minh \(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{1}{1+z^2}\ge\dfrac{9}{4}\Leftrightarrow1-\dfrac{1}{x^2}+1-\dfrac{1}{1+y^2}+1-\dfrac{1}{z^2+1}\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{x^2}{x^2+1}+\dfrac{y^2}{y^2+1}+\dfrac{z^2}{z^2+1}\ge\dfrac{3}{4}\)
mà \(\dfrac{x^2}{x^2+1}+\dfrac{y^2}{y^2+1}+\dfrac{z^2}{z^2+1}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3}=\dfrac{x^2+y^2+z^2+2}{x^2+y^2+z^2+3}=1-\dfrac{1}{x^2+y^2+z^2+3}\ge\dfrac{3}{4}\)
=> BĐT cầnd chứng minh luôn đúng
Cho \(x+y+z=0\)
Chứng minh rằng: \(a^5\left(b^2+c^2\right)+b^5\left(a^2+c^3\right)+c^5\left(a^2+b^2\right)=\dfrac{1}{2}\left(a^3+b^3+c^3\right)\left(a^4+b^4+c^4\right)\)
rút gọn các phân thức
a) \(\frac{x^2-16}{4x-x^2}\left(x\ne0,x\ne4\right)\) d) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ab}\)
b) \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}\left(x\ne y\right)\) c) \(\frac{\left(x+y\right)^2-z^2}{x+y+z}\left(x+y+z\ne0\right)\)
e)\(\frac{a^3+b^3+c^3}{a^2+b^2+c^2-ab-bc-ac}\)
CMR:
a,\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
b,\(\left(x+y+z\right)^2\ge3\cdot\left(xy+yz+xz\right)\)
