\(\frac{\left(a+b\right)^2}{4}\le\frac{a^2+b^2}{2}\Leftrightarrow4a^2+4b^2\ge2\left(a^2+2ab+b^2\right)=2a^2+4ab+2b^2\)

\(\Leftrightarrow2a^2-4ab+2b^2\ge0\Leftrightarrow2\left(a-b\right)^2\ge0\) (đúng)

Vậy ta có đpcm.

\(\sqrt{a^2+b^2}>=\dfrac{a+b}{\sqrt{2}}\)

=>\(\sqrt{2a^2+2b^2}>=a+b\)

=>\(2a^2+2b^2>=\left(a+b\right)^2=a^2+2ab+b^2\)

=>\(a^2-2ab+b^2>=0\)

=>\(\left(a-b\right)^2>=0\)(luôn đúng)

\(a^2+b^2+4\ge ab+2\left(a+b\right)\)

\(\Leftrightarrow2a^2+2b^2+8\ge2ab+4\left(a+b\right)\)

\(\Leftrightarrow2a^2+2b^2+8-2ab-4a-4b\ge0\)

\(\Leftrightarrow a^2-2ab+b^2+a^2-4a+4+b^2-4b+4\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-2\right)^2+\left(b-2\right)^2\ge0\) (Luôn đúng)

Vậy đẳng thức ban đầu được chứng minh.

\(a^2+b^2=a^2-2ab+b^2+2ab=\left(a-b\right)^2+2ab\)

Vì  \(\left(a-b\right)^2\ge0\Rightarrow\left(a-b\right)^2+2ab\ge2ab\left(dpcm\right)\)

\(\left(a+2\right)^2+\left(b+2\right)^2+\left(a^2+b^2+ab\right)\\ =a^2+4a+4+b^2+4b+4+a^2+b^2+ab\\ =2a^2+2b^2+4a+4b+ab+8\\ =\left[\left(a^2+ab+\dfrac{1}{4}b^2\right)+2\left(a+\dfrac{1}{2}b\right)+1\right]+\left(a^2+2a+1\right)+\dfrac{7}{4}\left(b^2+2\cdot\dfrac{6}{7}b+\dfrac{42}{49}\right)+\dfrac{9}{2}\\ =\left(a+\dfrac{1}{2}b+1\right)^2+\left(a+1\right)^2+\dfrac{7}{4}\left(b+\dfrac{6}{7}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}>0\left(đpcm\right)\)

Nó hot quá.2 giờ rồi câu hỏi đấy vẫn đứng ở đầu

Lời giải:

$a^2+b^2+1011-(ab+a+b)=\frac{2a^2+2b^2+2022-2ab-2a-2b}{2}$

$=\frac{(a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1)+2020}{2}$

$=\frac{(a-b)^2+(a-1)^2+(b-1)^2+2020}{2}$

$\geq \frac{2020}{2}>0$

$\Rightarrow a^2+b^2+1011> ab+a+b$

Ta có đpcm.