\(a^2+1\ge2a;b^2+1\ge2b;a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2+1+b^2+1+a^2+b^2\ge2a+2b+2ab\)\(\Leftrightarrow a^2+b^2+1\ge ab+a+b\left(đpcm\right)\)
Vậy \(a^2+b^2+1\ge ab+a+b\)
ta có
\(2\left(a^2+b^2+1\right)-2\left(ab+a+b\right)=2a^2+2b^2+2-2ab-2a-2b=\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\)với mọi a,b ta luôn có
\(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(a-1\right)^2\ge\\\left(b-1\right)^2\ge0\end{matrix}\right.0}\)=>\(\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)
<=>\(2\left(a^2+b^2+1\right)-2\left(ab+a+b\right)\ge0\Leftrightarrow a^2+b^2+1-ab -a-b\ge0\Leftrightarrow a^2+b^2+1\ge ab+a+b\)
`a^2+b^2+1geab+a+b`
Áp dụng BĐT cauchy cho hai số dương `a^2;b^2` ta có:
`a^2+b^2ge2sqrt{a^2b^2}=2ab`
Tương tự đối với `b^2;1` và `a^2;1`
`b^2+1ge2sqrt{b^2 . 1}=2b`
`a^2+1ge2sqrt{a^2 . 1}=2a`
`<=>a^2+b^2+b^2+1+a^2+1ge2ab+2b+2a`
`<=>2.(a^2+b^2+1)ge2.(ab+b+a)`
`<=>a^2+b^2+1geab+b+a(ĐPCM)`
Dấu `=` xảy ra khi:`a=b=1`