\(\Delta=m^2-4\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2\ge0\) ;\(\forall m\)

\(\Rightarrow\) Phương trình luôn có nghiệm với mọi m

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

\(a^4+1-a\left(a^2+1\right)=a^4+1-a^3-a=\left(a^4-a\right)-\left(a^3-1\right)\)

\(=a\left(a^3-1\right)-\left(a^3-1\right)=\left(a-1\right)\left(a^3-1\right)\)

\(=\left(a-1\right)\left(a-1\right)\left(a^2+a+1\right)=\left(a-1\right)^2\left[\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\right]\ge0\forall a\)(đpcm)

x²+y²+z²+3> hoac = 2(x+y+z)

\(x^2+y^2+z^2+3-2\left(x+y+z\right)\ge0\)

\(\Rightarrow x^2+y^2+z^2+3-2x-2y-2z\ge0\)

\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(z^2-2z+1\right)\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\ge0\)(Đpcm)

Dấu = khi (x-1)²=(y-1)²=(z-1)²=0 =>x=y=z=1

\(y'=x^2-2mx+m\)

\(y'\ge0\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta'\le0\end{matrix}\right.\Leftrightarrow m^2-m\le0\Leftrightarrow0\le m\le1\)

ta có (a+b+c)(1/a+1/b+1/c)=1+b/a+c/a+a/b+1+c/b+a/c+b/c+1=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)

ta có (a-b)²>0suy ra a/b+b/a> hoặc =2

suy ra (a+b+c)(1/a+1/b+1/c)>hoặc=9

suy ra 1/a+1/b+1/c>hoặc=9/a+b+c

1. Đ

2 S    ( lớn hơn hoặc =.)

3S    ( thêm hoặc =. vd x = 0)

4Đ

5S ( với mọi x >0)

6Đ

7Đ