cho a , b, c la cac so thuc duong thoa man he thuc a+b+c=6abc
Chung minh rang \(\dfrac{bc}{a^3\left(c+2b\right)}+\dfrac{ac}{b^3\left(a+2c\right)}+\dfrac{ab}{c^3\left(b+2a\right)}\ge2\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 7 2020
\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)
\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)
\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)
\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
24 tháng 11 2018
Ta có \(a+b+c+d=0\Leftrightarrow a+c=-\left(b+d\right)\Leftrightarrow\left(a+c\right)^3=\left[-\left(b+d\right)\right]^3\Leftrightarrow a^3+3a^2c+3ac^2+c^3=-b^3-3b^2d-3bd^2-d^3\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2c-3ac^2-3b^2d-3bd^2\Leftrightarrow a^3+b^3+c^3+d^3=-3ac\left(a+c\right)-3bd\left(b+d\right)\Leftrightarrow a^3+b^3+c^3+d^3=3ac\left(b+d\right)-3bd\left(b+d\right)\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)Vậy \(a+b+c+d=0\) thì \(a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 1 2022
\(\dfrac{a^3}{\left(a+2b\right)\left(b+2c\right)}+\dfrac{a+2b}{27}+\dfrac{b+2c}{27}\ge3\sqrt[3]{\dfrac{a^3\left(a+2b\right)\left(b+2c\right)}{27^2.\left(a+2b\right)\left(b+2c\right)}}=\dfrac{a}{3}\)
Tương tự:
\(\dfrac{b^3}{\left(b+2c\right)\left(c+2a\right)}+\dfrac{b+2c}{27}+\dfrac{c+2a}{27}\ge\dfrac{b}{3}\)
\(\dfrac{c^3}{\left(c+2a\right)\left(a+2b\right)}+\dfrac{c+2a}{27}+\dfrac{a+2b}{27}\ge\dfrac{c}{3}\)
Cộng vế:
\(VT+\dfrac{2\left(a+b+c\right)}{9}\ge\dfrac{a+b+c}{3}\)
\(\Rightarrow VT\ge\dfrac{a+b+c}{9}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
N
5 tháng 8 2017
từ giả thiết ,ta có:\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=4\)\(\Leftrightarrow a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1\)---> thay 1= vào ...
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 9 2019
\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=6\)
\(P=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}\)
\(P\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\frac{xy+yz+zx}{3}=2\)
Dấu "=" xảy ra khi \(x=y=z=\sqrt{2}\)
