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Ta có:
y02 + ay0 + b = 0
\(\Leftrightarrow\)y04 = (ay0 + b)2
\(\le\)(a2 + b2)(y02 + 1)
\(\Rightarrow\)y04 - 1 < (a2 + b2)(y02 + 1)
\(\Rightarrow\)y02 - 1 < a2 + b2
\(\Rightarrow\)y02 < 1 + a2 + b2
3/ Dễ thấy \(0\le x,y,z\le1\)
Ta có:
x2 + y2 + z2 = x3 + y3 + z3
\(\Leftrightarrow\)x2(1 - x) + y2(1 - y) + z2(1 - z) = 0
Dấu = xảy ra khi (x, y, z) = (0,0,1) và các hoán vị của nó
P=(2x+1/x)+(2y+1/y)-(x+y)+(x/y+y/x)+2
+có (x+y)^2 </ 2(x^2+y^2)(C-S) => x+y </ 2 => -(x+y) >/ căn (2)
+am-gm 3 lần
\(1>=\left(x+y\right)^2>=\left(2\sqrt{xy}\right)^2=4xy\Rightarrow1>=4xy\Rightarrow\frac{1}{2}>=2xy\)(bđt cosi)
\(\Rightarrow\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}>=\frac{4}{x^2+2xy+y^2}+\frac{1}{\frac{1}{2}}\)
\(=\frac{4}{\left(x+y\right)^2}+2>=\frac{4}{1^2}+2=4+2=6\)
dấu = xảy ra khi \(x=y=\frac{1}{2}\)
vậy min \(\frac{1}{x^2+y^2}+\frac{1}{xy}=6\)khi \(x=y=\frac{1}{2}\)
Áp dụng BDT AM-GM ta có:\(VT\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{x+y+1}\right)\)
\(\Rightarrow\frac{VT}{3}\ge\frac{x^2}{xy+xz+x}+\frac{y^2}{yz+yx+y}+\frac{z^2}{xz+zy+z}\)
\(\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+xy+z}\) (Cauchy-Schwarz)
Do \(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)\(\Rightarrow\left(x+y+z\right)^2\le\left(x^2+y^2+z^2\right)^2\)
\(\Rightarrow x+y+z\le x^2+y^2+z^2\).Suy ra
\(2\left(xy+yz+xz\right)+x+y+z\le2\left(xy+yz+xz\right)+x^2+y^2+z^2=\left(x+y+z\right)^2\)
Suy ra \(\frac{VT}{3}\le\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\Rightarrow VT\ge3\) (điều phải chứng minh)
Dấu "=" xảy ra khi x=y=z=1
Ta có:
\(A=\left(x^2+\frac{1}{8x}+\frac{1}{8x}\right)+\left(y^2+\frac{1}{8y}+\frac{1}{8y}\right)+\left(z^2+\frac{1}{8z}+\frac{1}{8z}\right)+\frac{6}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\ge3\sqrt[3]{x^2.\frac{1}{8x}.\frac{1}{8x}}+3\sqrt[3]{y^2.\frac{1}{8y}.\frac{1}{8y}}+3\sqrt[3]{z^2.\frac{1}{8z}.\frac{1}{8z}}+\frac{6}{8}\frac{9}{x+y+z}\)
\(=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{6}{8}.\frac{9}{\frac{3}{2}}=\frac{27}{4}\)
Dấu "=" xảy ra <=> x = y = z = 1/2
Vậy min A = 27/4 tại x = y = z = 1/2
\(VT=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\)
\(\ge\frac{3x}{y+z+1}+\frac{3y}{x+z+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+xz+x}+\frac{3y^2}{xy+yz+y}+\frac{3z^2}{xz+yz+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2}\)
\(\ge\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\ge xy+yz+xz=VP\)
Dấu "=" <=> x=y=z=1
2.
\(x+y+1=\sqrt{x}+\sqrt{y}+\sqrt{xy}\)
\(\Leftrightarrow2x+2y+2=2\sqrt{x}+2\sqrt{y}+2\sqrt{xy}\)
\(\Leftrightarrow x-2\sqrt{xy}+y+x-2\sqrt{x}+1+y-2\sqrt{y}+1=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=\sqrt{y}\\\sqrt{x}=1\\\sqrt{y}=1\end{matrix}\right.\Leftrightarrow x=y=1\)
Từ đó suy ra : \(\left\{{}\begin{matrix}P=1^2+1^2=2\\Q=1^{1023}+1^{2014}=2\end{matrix}\right.\)
1.
Xét \(x^3+y^3+xy=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)
\(=x^2-xy+y^2+xy\)( vì \(x+y=1\))
\(=x^2+y^2\)
Áp dụng bất đẳng thức Bunhiacopxki :
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2=1\)
\(\Rightarrow x^2+y^2\ge\frac{1}{2}\)
Từ đó ta có : \(P=\frac{1}{x^2+y^2}\le\frac{1}{\frac{1}{2}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)