Question

Với số m và số n bất kì, chứng tỏ rằng :

a) \(\left(m+1\right)^2\ge4m\)

b) \(m^2+n^2+2\ge2\left(m+n\right)\)

Answer 1

a. Ta có:

\(\left(m+1\right)^2\)\(=m^2+2m+1\)

\(\left(m+1\right)^2\ge4m\Leftrightarrow m^2+2m+1\ge4m\)

\(\Leftrightarrow m^2+2m+1-4m\ge0\)

\(\Leftrightarrow m^2-2m+1\ge0\)

\(\Leftrightarrow\left(m-1\right)^2\ge0\) (đúng \(\forall\) m)

Vậy \(\left(m+1\right)^2\ge4m\)

b. \(m^2+n^2+2\ge2\left(m+n\right)\)

\(\Leftrightarrow m^2+1+n^2+1\ge2m+2n\)

Ta có:

\(\left(m^2+1\right)^2\ge4m^2\) \(\Rightarrow m^2+1\ge2m\)

\(\left(n^2+1\right)^2\ge4n^2\Rightarrow n^2+1\ge2n\)