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\(A=2017+a^2+b^2+c^2\ge2017+\dfrac{1}{3}\left(a+b+c\right)^2=2020\)
\(A_{min}=2020\) khi \(a=b=c=1\)
\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Leftrightarrow4\ge\left(a+b\right)^2\Leftrightarrow-2\le a+b\le2\)
\(4ab\le2\left(a^2+b^2\right)\Leftrightarrow4ab\le4\Leftrightarrow ab\le1\)
\(A=\left(3-a\right)\left(3-b\right)=9-3a-3b+ab=9-3\left(a+b\right)+ab\le9+3.2+1=16\)
\(A_{max}=16\Leftrightarrow a=b=-1\)
\(\left(a+b\right)^2\ge2\left(a^2+b^2\right)=4\Rightarrow-2\le a+b\le2\)
\(P=9-3\left(a+b\right)+ab=9-3\left(a+b\right)+\dfrac{\left(a+b\right)^2-\left(a^2+b^2\right)}{2}\)
\(P=\dfrac{1}{2}\left(a+b\right)^2-3\left(a+b\right)+8\)
Đặt \(a+b=x\Rightarrow-2\le x\le2\)
\(P=\dfrac{1}{2}x^2-3x+8=\dfrac{1}{2}\left(x-2\right)\left(x-4\right)+4\)
Do \(-2\le x\le2\Rightarrow\left\{{}\begin{matrix}x-2\le0\\x-4< 0\end{matrix}\right.\) \(\Rightarrow\left(x-2\right)\left(x-4\right)\ge0\)
\(\Rightarrow P\ge4\Rightarrow P_{min}=4\) khi \(x=2\Leftrightarrow a=b=1\)
\(P=\dfrac{1}{2}\left(x+2\right)\left(x-8\right)+16\)
Do \(-2\le x\le2\Rightarrow\left\{{}\begin{matrix}x+2\ge0\\x-8< 0\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}\left(x+2\right)\left(x-8\right)\le0\)
\(\Rightarrow P\le16\Rightarrow P_{max}=16\) khi \(x=-2\Leftrightarrow a=b=-1\)
Với mọi số thực ta luôn có:
`(a-b)^2+(b-c)^2+(c-a)^2>=0`
`<=>a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2>=0`
`<=>2(a^2+b^2+c^2)>=2(ab+bc+ca)`
`<=>3(a^2+b^2+c^2)>=a^2+b^2+c^2+2(ab+bc+ca)`
`<=>3(a^2+b^2+c^2)>=(a+b+c)^2=4`
`<=>a^2+b^2+c^2>=4/3`
Dấu "=" xảy ra khi `a=b=c=2/3`
~Quang Anh Vũ~