Question

Chứng minh x^4+x^3+x^2+x+1>0

Answer 1

\(x^4+x^3+x^2+x+1=\left(x^4+x^3+\frac{1}{4}x^2\right)+\left(\frac{1}{4}x^2+x+1\right)+\frac{1}{2}x^2\)

\(=\left(x^2+\frac{1}{2}x\right)^2+\left(\frac{1}{2}x+1\right)^2+\frac{1}{2}x^2\ge0\) (Do từng hạng tử của đa thức đều \(\ge0\))

Nếu \(x=0\) thì

 \(\left(x^2+\frac{1}{2}x\right)^2+\left(\frac{1}{2}x+1\right)^2+\frac{1}{2}x^2=\left(0+\frac{1}{2}.0\right)^2+\left(\frac{1}{2}.0+1^2\right)+\frac{1}{2}.0^2=1>0\)

Do đó \(\left(x^2+\frac{1}{2}x\right)^2+\left(\frac{1}{2}x+1\right)^2+\frac{1}{2}x^2>0\) hay \(x^4+x^3+x^2+x+1>0\)