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\(P=\dfrac{1}{x^2+y^2+z^2}+\dfrac{2023}{xy+yz+zx}\)
\(=\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy+yz+zx}+\dfrac{1}{xy+yz+zx}+\dfrac{2021}{xy+yz+zx}\)
\(\ge\dfrac{9}{\left(x+y+z\right)^2}+\dfrac{2021}{\dfrac{\left(x+y+z\right)^2}{3}}\)\(=9+\dfrac{2021}{\dfrac{1}{3}}=6072\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Ta có:
+) \(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}\left(\text{Cô si}\right)\)
+) \(\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy+yz+zx}+\dfrac{1}{xy+yz+zx}\)
\(\ge\dfrac{9}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}=\dfrac{9}{\left(x+y+z\right)^2}\left(\text{Svácxơ}\right)\)
áp dụng BĐT C-S dạng engel : A >/ x+y+z
áp dụng BĐT AM-GM x+y+z >/ căn xy + căn yz + căn zx
=>minA = 1
Luôn có \(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-x\right)^2\ge0\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\ge0\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)\ge xy+yz+xz\ge-1\)
\(P_{min}=-1\)dấu "=" sảy ra khi (x,y,z) là hoán vị của 3 phần tử (0,0,-1)
Ta có:
\(xy+yz+zx=-1\)
\(\Leftrightarrow2\left(xy+yz+zx\right)=-2\)
\(\Leftrightarrow2\left(xy+yz+zx\right)+x^2+y^2+z^2=-2+x^2+y^2+z^2\)
\(\Leftrightarrow P=x^2+y^2+z^2=\left(x+y+z\right)^2+2\ge2\)
Dấu = xảy ra khi \(\hept{\begin{cases}x+y+z=0\\xy+yz+zx=-1\end{cases}}\)
Chỉ ra 1 bộ số thỏa mãn cái đấy nhé là: \(\hept{\begin{cases}x=0\\y=1\\z=-1\end{cases}}\)
1. Với mọi số thực x;y;z ta có:
\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)
\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)
\(\Rightarrow P\ge3\)
\(P_{min}=3\) khi \(x=y=z=1\)
1.1
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)
\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)
\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)
\(\Leftrightarrow a=b\Leftrightarrow x=y\)
Thay vào pt đầu:
\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))
\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)
\(\Rightarrow a=1\Rightarrow x=y=1\)
2.
\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)
\(\Rightarrow4x^2-10xy+4y^2=0\)
\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu
...
Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)
\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)
\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\)
(vì \(2013=3.671=3\left(xy+yz+zx\right)\))
\(=\dfrac{x+y+z}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}\)
\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)
\(=\dfrac{1}{x+y+z}\)
ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)
\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)
\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))
Vậy ta có đpcm.
Áp dụng BĐT AM-GM ta có:
\(\hept{\begin{cases}\sqrt{xy}\le\frac{x+y}{2}\\\sqrt{yz}\le\frac{y+z}{2}\\\sqrt{xz}\le\frac{x+z}{2}\end{cases}}\). Cộng theo vế ta có:
\(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=1\le\frac{x+y+y+z+x+z}{2}=\frac{2\left(x+y+z\right)}{2}=x+y+z\)
Do đó ta có: \(x+y+z\ge1\).Áp dụng BĐT Cauchy-Schwarz dạng Engel ta cũng có:
\(A\ge\frac{\left(x+y+z\right)^2}{x+y+y+z+x+z}=\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)