\(\Rightarrow(x^2+2x\dfrac{y}{2}+\dfrac{y^2}{4})+\dfrac{3y^2}{4}\)
\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}\)
Mà \(\left(x+\dfrac{y}{2}\right)^2\ge0;\dfrac{3y^2}{4}\ge0\)
\(\Rightarrow\)đpcm
\(\Rightarrow(x^2+2x\dfrac{y}{2}+\dfrac{y^2}{4})+\dfrac{3y^2}{4}\)
\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}\)
Mà \(\left(x+\dfrac{y}{2}\right)^2\ge0;\dfrac{3y^2}{4}\ge0\)
\(\Rightarrow\)đpcm
CMR: \(P=x^2+xy+y^2-3\left(x+y\right)+3\) có GTNN bằng 0
1.cho x+y+z=xyz và xy+yz+zx≠3
cmr: x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)/xy+yz+zx=xyz
2.cmr nếu c^2+2(ab-ac-bc)=0và b≠c,a+b≠c thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
3. cho a,b,c thỏa mãn abc≠0 và ab+bc+ca=0
tính :P=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
CMR: \(x^2+y^2+1\ge xy+x+y\)
cmr: 1/(1+x)^2+1/(1+y)^2>= 1/1+xy
1CMR: x2+y2+8\(\ge\) xy+2x+2y
2 Cho a+b+c=6 . Cmr: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{3}{4}\)
3 Cho x+y+z+xy+yz+zx=6. Cmr: x2+y2+z2 \(\ge3\)
Cho x,y > 0 ; x # y thỏa mãn 1/(1+x^2 + 1/(1+y^2) = 2/1+xy . Tính D= 1/(1+x^2) + 1/(1+y^2) + 1/(1 + xy)
Cho\(\dfrac{x^2-yx}{a}=\dfrac{y^2-xz}{b}=\dfrac{z^2-xy}{c}\).CMR:\(\dfrac{a^2-bc}{x}=\dfrac{b^2-ca}{y}=\dfrac{c^2-ab}{z}\)
a, Cho 0 <= x,y,z <= 1. Chứng minh
0 <= x+y+z-xy-yz-xz <=1
b, Cho -1 <= x,y,z <=2 và x+y+z=0 . Chứng minh
x^2 + y^2 + z^2 <= 6
Cho x,y,z > 0 và x+y+z+xy+yz+zx=6 .C/minh x^2 + y^2+ z^2 > hoặc = 6