\(!x+\frac{1}{x}!\ge2\Rightarrow!a!\ge2\\ \)

Với IaI>=2

ta có: \(\left(x+\frac{1}{x}\right)^2=a^2\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

Áp dụng bđt Cauchy Schwarz dạng Engel ta có:

\(\frac{a^2+b^2+c^2}{3}=\)(\(\frac{a^2}{1}+\frac{b^2}{1}+\frac{c^2}{1}\)).\(\frac{1}{3}\ge\)\(\frac{\left(a+b+c\right)^2}{1+1+1}.\frac{1}{3}=\)\(\left(\frac{a+b+c}{3}\right)^2\)(đpcm)

Dấu "=" xảy ra khi a = b = c

            Bài làm :

Áp dụng bất đẳng thức Cauchy Schwarz dạng Engel ta có:

\(\frac{a^2+b^2+c^2}{3}=\left(\frac{a^2}{1}+\frac{b^2}{1}+\frac{c^2}{1}\right).\frac{1}{3}\ge\frac{\left(a+b+c\right)^2}{1+1+1}.\frac{1}{3}=\left(\frac{a+b+c}{3}\right)^2\)

Dấu "=" xảy ra khi a = b = c

số còn lại là dư

91 điểm

95 điểm

\(\Leftrightarrow\left(x^4+x^3\right)+\left(2x^3+2x^2\right)+\left(2x^2+2x\right)+\left(x+1\right)\)\(=0\)

\(\Leftrightarrow x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+2x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)

\(Do\)\(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge0\)

\(\Rightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)