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9 tháng 12 2021
\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)
\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)
9 tháng 12 2021
\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
15 tháng 5 2018
\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{2}{c}=0\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{b}=\dfrac{2}{c}-\dfrac{1}{a}=\dfrac{2a-c}{ac}\\\dfrac{1}{a}=\dfrac{2}{c}-\dfrac{1}{b}=\dfrac{2b-c}{bc}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2a-c=\dfrac{ac}{b}\\2b-c=\dfrac{bc}{a}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+c}{2a-c}=\dfrac{b\left(a+c\right)}{ac}=\dfrac{ab}{ac}+\dfrac{bc}{ac}=\dfrac{b}{c}+\dfrac{b}{a}\\\dfrac{b+c}{2b-c}=\dfrac{a\left(b+c\right)}{bc}=\dfrac{ab}{bc}+\dfrac{ac}{bc}=\dfrac{a}{c}+\dfrac{a}{b}\end{matrix}\right.\)
Áp dụng bđt Cosi cho 2 số sương ta có: \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{2}{c}=0\Leftrightarrow\dfrac{2}{c}=\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Leftrightarrow\dfrac{a+b}{c}\ge2\)(áp dụng \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\))
Ta có: \(\dfrac{a+c}{2a-c}+\dfrac{b+c}{2b-c}=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\dfrac{a+b}{c}\ge2+2=4\)
Dấu "=" xawy ra khi và chỉ khi a=b=c
A
23 tháng 1 2018
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^2}{a+2b^2}+\dfrac{a+2b^2}{9}\ge2\sqrt{\dfrac{a^2}{a+2b^2}\cdot\dfrac{a+2b^2}{9}}=\dfrac{2a}{3}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT+\dfrac{a+b+c+2\left(a^2+b^2+c^2\right)}{9}\ge\dfrac{2}{3}\left(a+b+c\right)\)
\(\Leftrightarrow VT+\dfrac{3+2\cdot\dfrac{\left(a+b+c\right)^2}{3}}{9}\ge\dfrac{2}{3}\cdot3\)
\(\Leftrightarrow VT+1\ge2\Leftrightarrow VT\ge1\)
\("="\Leftrightarrow a=b=c=1\)
23 tháng 1 2018
WLOG \(a\ge b \ge c\)
Chebyshev: \(\left(a+b+c\right)\left(a^3+b^3+c^3\right)\le3\left(a^4+b^4+c^4\right)\)
\(\Rightarrow a^3+b^3+c^3\le a^4+b^4+c^4\)
Cauchy-Schwarz: \(VT=\dfrac{a^4}{a^3+2a^2b^2}+\dfrac{b^4}{b^3+2b^2c^2}+\dfrac{c^4}{c^3+2a^2c^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+2\left(a^2b^2+b^2c^2+c^2a^2\right)}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}\)
\(=\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2}=1=VP\)
YS
30 tháng 10 2018
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=-\dfrac{1}{c^3}\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}=0\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=0\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{3}{ab}.\left(-\dfrac{1}{c}\right)=0\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}-\dfrac{3}{abc}=0\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Ta có: Điều cần chứng minh là \(A=3abc\) hay \(\dfrac{A}{3abc}=1\)
Thật vậy:
\(\dfrac{A}{3abc}=\left(\dfrac{b^2c^2}{a}+\dfrac{c^2a^2}{b}+\dfrac{a^2b^2}{c}\right).\dfrac{1}{3abc}\)
\(\dfrac{A}{3abc}=\dfrac{b^2c^2}{3a^2bc}+\dfrac{c^2a^2}{3ab^2c}+\dfrac{a^2b^2}{3abc^2}\)
\(\dfrac{A}{3abc}=\dfrac{bc}{3a^2}+\dfrac{ac}{3b^2}+\dfrac{ab}{3c^2}\)
\(\dfrac{A}{3abc}=\dfrac{abc}{3a^3}+\dfrac{abc}{3b^3}+\dfrac{abc}{3c^3}\)
\(\dfrac{A}{3abc}=\dfrac{abc}{3}\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{abc}{3}.\dfrac{3}{abc}=1\)
\(\dfrac{A}{3abc}=1\Leftrightarrow A=3abc\left(đpcm\right)\)
5 tháng 11 2017
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{-1}{c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\dfrac{-1}{c^3}\) hay \(\dfrac{1}{a^3}+\dfrac{1}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{b^3}=\dfrac{-1}{c^3}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{3}{abc}\)
\(a^2b^2c^2.\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{3}{abc}.a^2b^2c^2\)
\(\Leftrightarrow\dfrac{b^2c^2}{a}+\dfrac{c^2a^2}{b}+\dfrac{a^2b^2}{c}=3abc\) hay\(M=3abc\left(đpcm\right)\)
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 \(a=b=c\)
Đặt \(\frac{2a}{b+c}=t\)(t>0).Ta cần c/m: \(t+\frac{1}{t}\ge2\) (t > 0)
Thật vậy,áp dụng BĐT AM-GM (Cô si),ta có \(t+\frac{1}{t}\ge2\sqrt{t.\frac{1}{t}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow t^2=1\Leftrightarrow t=1\Leftrightarrow2a=b+c\Leftrightarrow a=b=c\)