Cho a,b,c \(\in\)[-2,5] thỏa mãn a+2b+3c\(\le\)2.Chứng minh \(a^2+2b^2+3c\)\(\le\)66
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\(a\in\left[-2;5\right]\Rightarrow\left(a+2\right)\left(a-5\right)\le0\)
\(\Leftrightarrow a^2\le3a+10\)
Tương tự: \(b^2\le3b+10\Rightarrow2b^2\le6b+20\)
\(c^2\le3c+10\Rightarrow3c^2\le9c+30\)
Cộng vế:
\(a^2+2b^2+3c^2\le3\left(a+2b+3c\right)+60\le66\) (đpcm)
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Lời giải:
Vì \(a,b,c\in [-2;5]\) nên:
\(\left\{\begin{matrix} (a+2)(a-5)\leq 0\\ (b+2)(b-5)\leq 0\\ (c+2)(c-5)\leq 0\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} a^2\leq 3a+10\\ b^2\leq 3b+10\\ c^2\leq 3c+10\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} a^2\leq 3a+10\\ 2b^2\leq 6b+20\\ 3c^2\leq 9c+30\end{matrix}\right. \)
Do đó:
\(a^2+2b^2+3c^2\leq 3(a+2b+3c)+60\)
Mà \(a+2b+3c\leq 2\)
\(\Rightarrow a^2+2b^2+3c^2\leq 3.2+60=66\)
Ta có đpcm
Dấu bằng xảy ra khi \((a,b,c)=(-2,5,-2)\)
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Vì \(-2\le a;b;c\le5\Rightarrow\hept{\begin{cases}\left(a+2\right)\left(a-5\right)\le0\\2\left(b+2\right)\left(b-5\right)\le0\\3\left(c+2\right)\left(c-5\right)\le0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}a^2-3a-10\le0\\2b^2-6b-20\le0\\3c^2-9b-30\le0\end{cases}}\)
\(\Rightarrow a^2+2b^2+3c^2-3\left(a+2b+3c\right)-60\le0\)
\(\Rightarrow a^2+2b^2+3c^2\le3\left(a+2b+3c\right)+60\le3.2+60=66\) (ĐPCM)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}a=-2\\a=5\end{cases};\orbr{\begin{cases}b=-2\\b=5\end{cases};\orbr{\begin{cases}c=-2\\c=5\end{cases}}}}\)
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Điều đó là đương nhiên mà. Giả sử x2 + y2 + z2 = 5 thì x2 + y2 + z2 \(\le\) 5
Áp dụng bất đẳng thức Bu.nhia.cop.xki cho 2 bộ 3 số:
\(\left(a+2b+3c\right)^2=\left(1.a+\sqrt{2}.\sqrt{2}b+\sqrt{3}.\sqrt{3}c\right)^2\)
\(\le\left(1+2+3\right)\left(a^2+2b^2+3c^2\right)=6.6=36\)
\(\Rightarrow\left|a+2b+3c\right|\le6\)
\(\Rightarrow-6\le a+2b+3c\le6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{ab}{2b}\right)\)
\(=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Tương tự:
\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{b}{2}\right)\)
\(\dfrac{ac}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ac}{b+c}+\dfrac{ac}{a+b}+\dfrac{c}{2}\right)\)
Cộng vế:
\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{a+b+c}{2}\right)\)
\(P\le\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)