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Đặt \(A=\frac{xy\sqrt{z-1}+xz\sqrt{y-2}+yz\sqrt{x-3}}{xyz}\)
\(\Rightarrow A=\frac{\sqrt{z-1}}{z}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{x-3}}{x}\)
\(\Rightarrow A=\frac{2.\sqrt{z-1}}{2z}+\frac{2.\sqrt{2}.\sqrt{y-2}}{2.\sqrt{2}.y}+\frac{2.\sqrt{3}.\sqrt{x-3}}{2.\sqrt{3}.x}\)\
\(\Rightarrow A\le\frac{z-1+1}{2z}+\frac{y-2+2}{2\sqrt{2}.y}+\frac{z-3+3}{2\sqrt{3}.x}\) ( ÁP DỤNG BĐT CÔ-SI )
\(\Rightarrow A\le\frac{z}{2z}+\frac{y}{2\sqrt{2}.y}+\frac{z}{2\sqrt{3}.z}\)
\(\Rightarrow A\le\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}=\frac{1}{2}+\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{6}\)
5(x+y)2+3(x-y)2=8x2+4xy+8y2=4(2x2+xy+2z2)>=5(x+y)2
=> \(\sqrt{2x^2+xy+2y^2}\ge\sqrt{\frac{5\left(x+y\right)^2}{4}}\)= \(\frac{\sqrt{5}\left(x+y\right)}{2}\)
Tương tự. Cộng lại là ra nha. Dấu = xảy ra <=> x=y=z=1/3
\(\Leftrightarrow\sqrt{4x^2+4xy+8y^2}+\sqrt{4y^2+4yz+8z^2}+\sqrt{4z^2+4zx+8x^2}\ge4\left(x+y+z\right)\)
Ta có:
\(VT=\sqrt{\left(2x+y\right)^2+\left(\sqrt{7}y\right)^2}+\sqrt{\left(2y+z\right)^2+\left(\sqrt{7}z\right)^2}+\sqrt{\left(2z+x\right)^2+\left(\sqrt{7}x\right)^2}\)
\(VT\ge\sqrt{\left(2x+y+2y+z+2z+x\right)^2+\left(\sqrt{7}x+\sqrt{7}y+\sqrt{7}z\right)^2}\)
\(VT\ge\sqrt{16\left(x+y+z\right)^2}=4\left(x+y+z\right)\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\)
BĐT Mincopxki:
\(\sqrt{x^2+a^2}+\sqrt{y^2+b^2}+\sqrt{z^2+c^2}\ge\sqrt{\left(x+y+z\right)^2+\left(a+b+c\right)^2}\)
Cộng vế với vế của 3 đẳng thức đã cho ta được:
\(x+y+z-2\sqrt{y+2012}-2\sqrt{z-2013}-2\sqrt{x-2}=0\)
\(\Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y+2012-2\sqrt{y+2012}+1\right)+\left(z-2013+2\sqrt{z-2013}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2012}-1\right)^2+\left(\sqrt{z-2013}-1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(\sqrt{x-2}-1\right)^2=0\\\left(\sqrt{y+2012}-1\right)^2=0\\\left(\sqrt{z-2013}-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y+2012}-1=0\\\sqrt{z-2013}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y+2012}=1\\\sqrt{z-2013}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2011\\z=2014\end{matrix}\right.\)
Thay vào C ta được:
C = (3 - 4)2016 + (-2011 + 2012)2017 + (2014 - 2013)2018
C = 1 + 1 + 1 = 3
THÊM
Cho x, y, z thõa mãn đồng thời: \(3x-2y-2\sqrt{y+2012}+1=0;3y-2z-2\sqrt{z-2013}+1=0;3z-2x-2\sqrt{x-2-2=0.}\)Tính \(C=\left(x-4\right)^{2016}+\left(y+2012\right)^{2017}+\left(z-2013\right)^{2018}\)
Áp dụng BĐT Cô - si cho 3 số không âm:
\(1+x^3+y^3\ge3\sqrt[3]{1.x^3y^3}=3xy\Rightarrow\frac{\sqrt{1+x^3+y^3}}{xy}\ge\frac{\sqrt{3}}{\sqrt{xy}}\)
Tương tự ta có: \(\frac{\sqrt{1+y^3+z^3}}{yz}\ge\frac{\sqrt{3}}{\sqrt{yz}}\);\(\frac{\sqrt{1+z^3+x^3}}{zx}\ge\frac{\sqrt{3}}{\sqrt{zx}}\)
Cộng các vế của các BĐT trên, ta được:
\(\frac{\sqrt{1+x^3+y^3}}{xy}\)\(+\frac{\sqrt{1+y^3+z^3}}{yz}\)\(+\frac{\sqrt{1+z^3+x^3}}{zx}\ge\)\(\frac{\sqrt{3}}{\sqrt{xy}}\)\(+\frac{\sqrt{3}}{\sqrt{yz}}\)\(+\frac{\sqrt{3}}{\sqrt{zx}}\)
Tiếp tục áp dụng Cô - si:
\(BĐT\ge3\sqrt[3]{\frac{\sqrt{3}}{\sqrt{xy}}.\frac{\sqrt{3}}{\sqrt{yz}}.\frac{\sqrt{3}}{\sqrt{zx}}}=3\sqrt{3}\)
Vậy \(\frac{\sqrt{1+x^3+y^3}}{xy}\)\(+\frac{\sqrt{1+y^3+z^3}}{yz}\)\(+\frac{\sqrt{1+z^3+x^3}}{zx}\ge3\sqrt{3}\)
(Dấu "="\(\Leftrightarrow x=y=z=1\))
\(x^3+y^3+1=x^3+y^3+xyz\ge xy\left(x+y\right)+xyz=xy\left(x+y+z\right)\)
Tương tự:
\(y^3+z^3+1\ge yz\left(x+y+z\right);z^3+x^3+1\ge zx\left(x+y+z\right)\)
\(\Rightarrow VT\ge\frac{\sqrt{xy\left(x+y+z\right)}}{xy}+\frac{\sqrt{yz\left(x+y+z\right)}}{yz}+\frac{\sqrt{zx\left(x+y+z\right)}}{zx}\)
\(=\sqrt{x+y+z}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\right)\)
\(\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\frac{1}{\sqrt{xy}\cdot\sqrt{yz}\cdot\sqrt{zx}}}=3\sqrt{3}\)
Dấu "=" xảy ra tại \(x=y=z=1\)
