Question

Chứng minh rằng với a,b là hai số tuỳ ý ta có:

a² + b² +1 \(\ge\) ab + a +b

Answer 1

\(a^2+b^2+1\ge ab+a+b\)

\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)

\(\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b\ge0\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(a^2-2ab+b^2\right)\ge0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(a-b\right)^2\ge0\forall a,b\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}a-1=0\\b-1=0\\a-b=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\\a=b\end{matrix}\right.\Rightarrow a=b=1\)