Question

Chứng minh rằng: \(x^2+ 2x+1+y^2 +1 +2y \ge 2(x+1)(y+1)\)

Answer 1

Ta có:

\(\left[\left(x+1\right)-\left(y+1\right)\right]^2\ge0\) với mọi x,y

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

\(\Leftrightarrow\left(x+1\right)^2+\left(y+1\right)^2\ge2\left(x+1\right)\left(y+1\right)\)

\(\Leftrightarrow x^2+2x+1+y^2+2y+1\ge2\left(x+1\right)\left(y+1\right)\)

Vậy..

Answer 2

giả sử \(x^2+2x+1+y^2+1+2y\ge2\left(x+1\right)\left(y+1\right)\) đúng.

tương đương :\(\left(x+1\right)^2+\left(y+1\right)^2\ge2\left(x+1\right)\left(y+1\right)\)

\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\ge0\\ \Leftrightarrow\left[\left(x+1\right)-\left(y-1\right)\right]^2\ge0\left(\text{luôn đúng}\right)\)

nên \(x^2+2x+1+y^2+1+2y\ge2\left(x+1\right)\left(y+1\right)\) cũng đúng.