1: Cho x,y,z>0. CMR: \(\dfrac{x}{2x+y+z}+\dfrac{y}{x+2y+z}+\dfrac{z}{x+y+2z}\)
2: Cho 0<x<\(\dfrac{1}{2}\). CMR: \(\dfrac{1}{x}+\dfrac{2}{1+2x}\ge8\\\)
3: Cho x,y>0 và x+y=1. CMR:
a)\(\dfrac{1}{xy}+\dfrac{2}{x^2+y^2}\ge8\)
b)\(\dfrac{1}{xy}+\dfrac{1}{x^2+y^2}\ge6\\
\)
4: CM các bđt sau: a) \(x^3+4x+13x^2\)
b)\(x^4-x+\dfrac{1}{2}0\)
5: Cho a,b,c là độ dài 3 cạnh 1 tam giác. CMR:...
Đọc tiếp
1: Cho x,y,z>0. CMR: \(\dfrac{x}{2x+y+z}+\dfrac{y}{x+2y+z}+\dfrac{z}{x+y+2z}\)
2: Cho 0<x<\(\dfrac{1}{2}\). CMR: \(\dfrac{1}{x}+\dfrac{2}{1+2x}\ge8\\\)
3: Cho x,y>0 và x+y=1. CMR:
a)\(\dfrac{1}{xy}+\dfrac{2}{x^2+y^2}\ge8\)
b)\(\dfrac{1}{xy}+\dfrac{1}{x^2+y^2}\ge6\\ \)
4: CM các bđt sau: a) \(x^3+4x+1>3x^2\)
b)\(x^4-x+\dfrac{1}{2}>0\)
5: Cho a,b,c là độ dài 3 cạnh 1 tam giác. CMR:
a)\(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\)
b)\(\dfrac{1}{a+b},\dfrac{1}{b+c},\dfrac{1}{c+a}\)là 3 cạnh của 1 tam giác(cần CM theo bđt tam giác)
6: Cho a,b,c,d>0 và abcd=1. CMR:
\(a^2+b^2+c^2+d^2+ab+cd\ge6\)
5) a) Đặt b+c-a=x;a+c-b=y;a+b-c=z thì 2a=y+z;2b=x+z;2c=x+y
Ta có:
\(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)
Vậy ta suy ra đpcm
b) Ta có: a+b>c;b+c>a;a+c>b
Xét: \(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)
.Tương tự:
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c};\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)
Vậy ta có đpcm
6) Ta có:
\(a^2+b^2+c^2+d^2+ab+cd\ge2ab+2cd+ab+cd=3\left(ab+cd\right)\)
\(ab+cd=ab+\dfrac{1}{ab}\ge2\)
Suy ra đpcm