Cho các số thực x, y, z thỏa mãn |x|-3=|y|+4=10-|z|
Tính GTLN của K=y(x+z)
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Ta có P \(\le\dfrac{1^2+\left(\sqrt{x-1}\right)^2}{2}+\dfrac{2^2+\left(\sqrt{y-4}\right)^2}{2}+\dfrac{3^2+\left(\sqrt{z-9}\right)^2}{2}\)
\(=\dfrac{1+x-1+4+y-4+9+z-9}{2}=\dfrac{x+y+z}{2}=\dfrac{28}{2}=14\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}1=\sqrt{x-1}\\2=\sqrt{y-4}\\3=\sqrt{z-9}\end{matrix}\right.\Leftrightarrow x=2;y=8;z=18\)(tm)
\(2=x^2+y^2+z^2\ge y^2+z^2\ge2yz\Rightarrow yz\le1\)
\(P=x\left(1-yz\right)+y+z\Rightarrow P^2\le\left[x^2+\left(y+z\right)^2\right]\left[\left(1-yz\right)^2+1\right]\)
\(P^2\le\left(2+2yz\right)\left(y^2z^2-2yz+2\right)\)
\(P^2\le2\left(yz\right)^3-2\left(yz\right)^2+4=2y^2z^2\left(yz-1\right)+4\le4\)
\(\Rightarrow P\le2\)
\(P_{max}=2\) khi \(\left(x;y;z\right)=\left(0;1;1\right)\) và các hoán vị
\(P=x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^3=\dfrac{64}{3}\)
\(P_{min}=\dfrac{64}{3}\) khi \(x=y=z=\dfrac{4}{3}\)
Đặt \(\left(x;y;z\right)=\left(a+1;b+1;c+1\right)\Rightarrow\left\{{}\begin{matrix}a+b+c=1\\a;b;c\ge0\end{matrix}\right.\)
\(\Rightarrow0\le a;b;c\le1\) \(\Rightarrow\left\{{}\begin{matrix}a^2\le a\\b^2\le b\\c^2\le c\end{matrix}\right.\) \(\Rightarrow a^2+b^2+c^2\le a+b+c=1\)
\(P=\left(a+1\right)^2+\left(b+1\right)^2+\left(c+1\right)^2\)
\(P=a^2+b^2+c^2+2\left(a+b+c\right)+3=a^2+b^2+c^2+5\le1+5=6\)
\(P_{max}=6\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị hay \(\left(x;y;z\right)=\left(1;1;2\right)\) và hoán vị
Đặt \(P=xyz\le\dfrac{1}{4}\left(x+y\right)^2z=\dfrac{1}{4}\left(x+y\right)^2\left(2016-x-y\right)\)
Do \(\left\{{}\begin{matrix}x\ge2\\y\ge9\\z\ge1951\\x+y=2016-z\end{matrix}\right.\) \(\Rightarrow11\le x+y\le65\)
Đặt \(x+y=a\Rightarrow11\le a\le65\)
\(4P\le a^2\left(2016-a\right)=-a^3+2016a^2-8242975+8242975\)
\(4P\le\left(65-a\right)\left[\left(a^2-65^2\right)-1951\left(a-11\right)-144051\right]+8242975\le8242975\)
\(\Rightarrow P\le\dfrac{8242975}{4}\)
Dấu "=" xảy ra khi \(\left[{}\begin{matrix}x=y=\dfrac{65}{2}\\z=1951\end{matrix}\right.\)
\(A=xy+xz+2yz+2xz=x\left(y+z\right)+2z\left(x+y\right)\)
\(=x\left(6-x\right)+2z\left(6-z\right)=-x^2+6x+2\left(-z^2+6z\right)\)
\(=-\left(x-3\right)^2-2\left(z-3\right)^2+27\le27\)
\(A_{max}=27\) khi \(\left(x;y;z\right)=\left(3;0;3\right)\)