Cho \(x^2+y^2=2\)
\(CM:P=xy.\left(x-y\right)^2\le\frac{1}{2}\)
K
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26 tháng 4 2020
Ta có \(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\frac{\left(yz+1\right)^2}{z^2}}{\frac{zx+1}{x}}+\frac{\frac{\left(zx+1\right)^2}{x^2}}{\frac{xy+1}{y}}+\frac{\frac{\left(xy+1\right)^2}{y^2}}{\frac{yz+1}{z}}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT \(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\frac{a_3^2}{b_3}\ge\frac{\left(a_1+a_2+a_3\right)^2}{b_1+b_2+b_3}\)
Dấu "=" xảy ra khi \(\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{c_3}\)
\(P=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\)
\(P\ge a+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Áp dụng BĐT: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
=> \(P\ge x+y+z+\frac{9}{x+y+z}=\left[x+y+z+\frac{9}{4\left(x+y+z\right)}\right]+\frac{27}{4\left(x+y+z\right)}\)
Ta có: \(x+y+z+\frac{9}{4\left(x+y+z\right)}\ge2\sqrt{\frac{9}{4}}=3;\frac{27}{4\left(x+y+z\right)}=\frac{27}{4\cdot\frac{3}{2}}=\frac{9}{2}\)
=> \(P\ge3+\frac{9}{2}=\frac{15}{2}\).
Dấu "=" xảy ra <=> x=y=z=\(\frac{1}{2}\)
Vậy MinP=\(\frac{15}{2}\)đạt được khi x=y=z=\(\frac{1}{2}\)
KN
26 tháng 4 2020
Ta có:
\(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\left(\frac{yz+1}{z}\right)^2}{\left(\frac{zx+1}{x}\right)}+\frac{\left(\frac{zx+1}{x}\right)^2}{\left(\frac{xy+1}{y}\right)}+\frac{\left(\frac{xy+1}{y}\right)^2}{\left(\frac{yz+1}{z}\right)}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT Bunhiacopxki dạng phân thức, ta có:
\(\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)\(\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\ge\left(x+y+z\right)+\frac{9}{x+y+z}=\left(x+y+z\right)+\frac{9}{4\left(x+y+z\right)}\)
\(+\frac{27}{4\left(x+y+z\right)}\ge2\sqrt{\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}}+\frac{27}{4.\frac{3}{2}}=\frac{15}{2}\)(Áp dụng BĐT Cô - si cho 2 số không âm)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 6 2020
\(\frac{3}{2}\ge x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(P\ge3\sqrt[3]{\frac{x\left(yz+1\right)^2.y\left(zx+1\right)^2.z\left(xy+1\right)^2}{z^2\left(zx+1\right)x^2\left(xy+1\right)y^2\left(yz+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\)
Xét \(Q=\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}=\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{\sqrt{xy}.\sqrt{yz}.\sqrt{zx}}\)
Đặt \(\left(\sqrt{xy};\sqrt{yz};\sqrt{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c\le\frac{3}{2}\Rightarrow abc\le\frac{1}{8}\)
\(Q=\frac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}=\frac{1+a^2b^2c^2+a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2}{abc}\)
\(Q\ge\frac{1+a^2b^2c^2+3\sqrt[3]{a^2b^2c^2}+3\sqrt[3]{a^4b^4c^4}}{abc}=\frac{1}{abc}+abc+3\left(\frac{1}{\sqrt[3]{abc}}+\sqrt[3]{abc}\right)\)
\(Q\ge abc+\frac{1}{64abc}+3\left(\sqrt[3]{abc}+\frac{1}{4\sqrt[3]{abc}}\right)+\frac{63}{64abc}+\frac{9}{4\sqrt[3]{abc}}\)
\(Q\ge2\sqrt{\frac{abc}{64abc}}+6\sqrt{\frac{\sqrt[3]{abc}}{4\sqrt[3]{abc}}}+\frac{63}{64.\frac{1}{8}}+\frac{9}{4.\sqrt[3]{\frac{1}{8}}}=\frac{125}{8}\)
\(\Rightarrow P\ge3\sqrt[3]{Q}\ge3\sqrt[3]{\frac{125}{8}}=\frac{15}{2}\)
\(P_{min}=\frac{15}{2}\) khi \(a=b=c=\frac{1}{2}\) hay \(x=y=z=\frac{1}{2}\)
13 tháng 7 2019
A
Áp dụng BĐT cosi ta có
\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)
\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)
Khi đó
\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)
MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)
13 tháng 7 2019
B
Áp dụng BĐT cosi ta có :
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)
Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)
=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)
\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z
24 tháng 3 2020
Theo bài ra ta có: \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\Rightarrow x+y+z=xyz\)
Do:\(\sqrt{yz\left(1+x^2\right)}=\sqrt{yz+x^2yz}=\sqrt{yz+x\left(x+y+z\right)}=\sqrt{\left(x+y\right)\left(x+z\right)}\)
Tương tự: \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(x+z\right)}\);
\(\sqrt{zx\left(1+y^2\right)}=\sqrt{\left(z+y\right)\left(x+y\right)}\)
\(A=\sqrt{\frac{x^2}{yz\left(1+x^2\right)}}+\sqrt{\frac{y^2}{zx\left(1+y^2\right)}}+\sqrt{\frac{z^2}{xy\left(1+z^2\right)}}\)
\(A=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\)
Áp dụng bất đẳng thức Cô si \(\frac{a+b}{2}\ge\sqrt{ab}\), dấu "=" xảy ra khi \(a=b\)
Ta có \(\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\);
\(\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)\);
\(\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\le\frac{1}{2}\left(\frac{z}{x+z}+\frac{z}{y+z}\right)\)
\(A\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+z}+\frac{y}{y+x}+\frac{z}{y+z}+\frac{z}{x+z}\right)=\frac{3}{2}\)
Vậy \(A\le\frac{3}{2}\). Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)
DH
24 tháng 3 2020
M giải thích cho t chỗ sao mà \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(x+z\right)}\) đc vậy?
Với cả từ dòng này xuống dòng này nữa.
Sao mà tin đc dấu " = " xảy ra khi nào vậy?
Cách làm tương tự câu trên:
Xét 2.P - 1 = 2xy. (x2 + y2 - 2xy) - 1 = 2xy.(2 - 2xy) - 1 = -4(xy)2 + 4xy - 1 = - (2xy - 1)2 \(\le\) 0 với mọi x; y
=> 2P - 1 \(\le\) 0 => P \(\le\) 1/2
Dấu "=" xảy ra <=> 2xy = 1 => (x -y)2 = 1 => x - y = 1 hoặc x - y = -1
mà (x+y)2 = x2 + y2 + 2xy = 3 => x + y = \(\sqrt{3}\) hoặc x + y = - \(\sqrt{3}\)
+) x - y = 1; x + y = \(\sqrt{3}\) => x = (\(\sqrt{3}\) + 1)/2 ; y = (\(\sqrt{3}\) - 1)/2
Hoặc x - y = 1; x+ y = - \(\sqrt{3}\): làm tương tự.....