Xét hiệu: \(x^5+y^5-x^4y-xy^4=x^4\left(x-y\right)-y^4\left(x-y\right)\)
\(=\left(x-y\right)\left(x^4-y^4\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\)≥ 0. Dấu "=" xảy ra khi x=y
Vậy x5+y5 ≥ x4y+xy4. Dấu "=" xảy ra khi x=y
c/m voi moi so thuc x,y,z ta co
x2+y2+z2+3> hoac = 2(x+y+z)
c/m
x+1/x> hoac=2 voi moi x>0
c/m voi moi a,b,c>0 ta co
1/a+1/b+1/c> hoac =9/a+b+c
chung minh P=x^4 - 2x^3 + 2x^2 - 2x + 1 lon hon hoac = 0 voi moi gia tri cua x
voi moi so a b c tuy y. CMR: a^2 + b^2 + c^2 > hoac = 2ab - 2ac + 2bc
Chung Minh a^4+1-a(a^2+1)>hoac = 0 voi moi a
c/m \(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\)
voi moi x,y>0
c/m voi 3 so thuc tuy y ta co
a2 cong b2 cong 1 lon hon hoac bang ab cong a cong b
2/ Chung to: x^2+6x+13>0 voi moi x
3/ Chung to : -x^2+6x-11<0 voi moi x
4/ cho x-y=5va x.y=24 .Tinh x^3+y^3
