Cho \(\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\left(x,y,z\ne0\right)\)
Tìm: \(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
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PN
4 tháng 8 2020
Bạn tham khảo câu trả lời của anh Phan Thanh Tịnh nhé
vô phần thống kê hỏi đáp của mình để coi hình nhé
5 tháng 8 2020
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\left(x^2-yz\right)\left(y-xyz\right)=\left(y^2-xz\right)\left(x-xyz\right)\)
\(\Leftrightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+xy^3z+x^2z-x^2yz^2=0\)
\(\Leftrightarrow xy\left(x-y\right)-xyz\left(x^2-y^2\right)+z\left(x^2-y^2\right)-xyz^2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[xy-xyz\left(x+y\right)+z\left(x+y\right)-xyz^2\right]=0\)
\(\Leftrightarrow xy-xyz\left(x+y\right)+z\left(x+y\right)-xyz^2=0\left(x\ne y\Rightarrow x-y\ne0\right)\)
\(\Leftrightarrow xy+yz+xz=xyz\left(x+y\right)+xyz^2\)
\(\Leftrightarrow\frac{ay+yz+xz}{xyz}=\frac{xyz\left(x+y\right)+xyz^2}{xyz}\left(xyz\ne0\right)\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=x+y+z\)
6 tháng 10 2019
Áp dụng t/c dãy tỉ số bằng nhau có:
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}=\frac{x^2-yz-y^2+xz}{x-xyz-y\left(1-xz\right)}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)
=> \(\frac{x^2-yz}{x\left(1-yz\right)}=x+y+z\)
<=> \(\frac{x^2-yz}{x\left(1-yz\right)}-\frac{\left(x+y+z\right)x\left(1-yz\right)}{x\left(1-yz\right)}=0\)
<=> \(\frac{x^2-yz-\left(x^2+yx+zx\right)\left(1-yz\right)}{x\left(1-yz\right)}\)=0
<=> \(x^2-yz-x^2+x^2yz-xy+xy^2z-xz+xyz^2=0\)
<=> \(-yz-xy-xz+xyz\left(x+y+z\right)\)=0
<=> \(xyz\left(x+y+z\right)=yz+xy+xz\)
<=>\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)( chia cả hai vế cho xyz với x,y,z khác 0)
CT
8 tháng 12 2019
vt sai đề nâk
từ gt=> xy+yz+xz=0
áp dụng bdt bunhia
=> A>=0
dấu= xr khi x=y=z
-> dấu = k xr
..........
hoặc:
8 tháng 12 2019
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\frac{\Rightarrow1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
\(\Rightarrow\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)
\(\Rightarrow xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)
RR
14 tháng 5 2018
Từ dữ kiện đề bài => x + y + z = xyz
Ta có :
\(\frac{x}{\sqrt{yz\left(1+x^2\right)}}=\frac{x}{\sqrt{yz+xyz.x}}=\frac{x}{\sqrt{yz+x\left(x+y+z\right)}}=\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}\)
\(=\frac{\sqrt{x}}{\sqrt{x+z}}.\frac{\sqrt{x}}{\sqrt{x+y}}\le\frac{1}{2}.\left(\frac{x}{x+z}+\frac{x}{x+y}\right)\)
Tương tự với hai hạng tử còn lại , suy ra
\(Q\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{x}{x+y}\right)+\frac{1}{2}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)+\frac{1}{2}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)
Vậy Max = 3/2 <=> x = y = z
Nguồn : Đinh Đức Hùng
KK
21 tháng 1 2017
Áp dụng BĐT Cô - si cho 3 bộ số không âm
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)
\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)
\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)
Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)
Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
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 \(x=y=z=\frac{1}{2}\)
4 tháng 1 2020
Đặt \(a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}\Rightarrow\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=1\end{matrix}\right.\)
\(K=\frac{\frac{1}{a}}{\sqrt{\frac{1}{bc}\left(1+\frac{1}{a^2}\right)}}+\frac{\frac{1}{b}}{\sqrt{\frac{1}{ac}\left(1+\frac{1}{b^2}\right)}}+\frac{\frac{1}{c}}{\sqrt{\frac{1}{ab}\left(1+\frac{1}{c^2}\right)}}\) \(=\frac{\frac{1}{a}}{\sqrt{\frac{a^2+1}{a^2bc}}}+\frac{\frac{1}{b}}{\sqrt{\frac{b^2+1}{ab^2c}}}+\frac{\frac{1}{c}}{\sqrt{\frac{c^2+1}{abc^2}}}\)
\(=\sqrt{\frac{bc}{a^2+1}}+\sqrt{\frac{ca}{b^2+1}}+\sqrt{\frac{ab}{c^2+1}}\) \(=\sqrt{\frac{bc}{a^2+ab+bc+ca}}+\sqrt{\frac{ca}{b^2+ab+bc+ca}}+\sqrt{\frac{ab}{c^2+ab+bc+ca}}\)
\(=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\)
\(\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}+\frac{a}{a+b}+\frac{c}{b+c}+\frac{a}{a+c}+\frac{b}{b+c}\right)\) \(\Rightarrow K\le\frac{3}{2}\)
Dấu "=" \(\Leftrightarrow a=b=c\Leftrightarrow x=y=z=\sqrt{3}\)
