LV
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
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: \(VT=\dfrac{a^4}{a^3+2a^2b^2}+\dfrac{b^4}{b^3+2b^2c^2}+\dfrac{c^4}{c^3+2a^2c^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+2\left(a^2b^2+b^2c^2+c^2a^2\right)}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}\)
\(=\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2}=1=VP\)
KK
12 tháng 5 2017
a) Áp dụng bất đẳng thức Schur với \(r=1\)
\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)
\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)
\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
b) Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
c) Ta có \(abc=ab+bc+ca\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)
\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\dfrac{3}{16}\)
\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )
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:
\(\dfrac{A}{3abc}=\left(\dfrac{b^2c^2}{a}+\dfrac{c^2a^2}{b}+\dfrac{a^2b^2}{c}\right).\dfrac{1}{3abc}\)
\(\dfrac{A}{3abc}=\dfrac{b^2c^2}{3a^2bc}+\dfrac{c^2a^2}{3ab^2c}+\dfrac{a^2b^2}{3abc^2}\)
\(\dfrac{A}{3abc}=\dfrac{bc}{3a^2}+\dfrac{ac}{3b^2}+\dfrac{ab}{3c^2}\)
\(\dfrac{A}{3abc}=\dfrac{abc}{3a^3}+\dfrac{abc}{3b^3}+\dfrac{abc}{3c^3}\)
\(\dfrac{A}{3abc}=\dfrac{abc}{3}\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{abc}{3}.\dfrac{3}{abc}=1\)
\(\dfrac{A}{3abc}=1\Leftrightarrow A=3abc\left(đpcm\right)\)
9 tháng 4 2017
\(\dfrac{b+c-a}{2a}+\dfrac{a-b+c}{2b}+\dfrac{a+b-c}{2c}\ge\dfrac{3}{2}\)
Ta có: \(\dfrac{b+c-a}{2a}=\dfrac{b}{2a}+\dfrac{c}{2a}-\dfrac{a}{2a}=\dfrac{b}{2a}+\dfrac{c}{2a}-\dfrac{1}{2}\)
Viết lại BĐT cần chứng minh như sau:
\(\dfrac{b}{2a}+\dfrac{c}{2a}-\dfrac{1}{2}+\dfrac{a}{2b}-\dfrac{1}{2}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-\dfrac{1}{2}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-\dfrac{3}{2}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-3\ge0\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{b}{2a}+\dfrac{a}{2b}=\dfrac{1}{2}\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{b}{a}\cdot\dfrac{a}{b}}=2\cdot\dfrac{1}{2}=1\)
\(\dfrac{c}{2a}+\dfrac{a}{2c}=\dfrac{1}{2}\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{c}{a}+\dfrac{a}{c}}=\dfrac{1}{2}\cdot2=1\)
\(\dfrac{b}{2c}+\dfrac{c}{2b}=\dfrac{1}{2}\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{b}{c}\cdot\dfrac{c}{b}}=\dfrac{1}{2}\cdot2=1\)
Cộng theo vế 3 BĐT trên ta có:
\(\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}\ge3\)
\(\Rightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-3\ge3-3=0\)
BĐT đúng nên ta có ĐPCM
AH
Akai Haruma
Giáo viên
9 tháng 4 2018
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(B=\frac{1}{(a+2b)(a+2c)}+\frac{1}{(b+2a)(b+2c)}+\frac{1}{(c+2a)(c+2b)}\)
\(\geq \frac{9}{(a+2b)(a+2c)+(b+2a)(b+2c)+(c+2a)(c+2b)}\)
\(\Leftrightarrow B\geq \frac{9}{(a^2+2ac+2ab+4bc)+(b^2+2bc+2ab+4ac)+(c^2+2bc+2ac+4ab)}\)
\(\Leftrightarrow B\geq \frac{9}{a^2+b^2+c^2+8(ab+bc+ac)}=\frac{9}{(a+b+c)^2+6(ab+bc+ac)}(*)\)
Theo hệ quả quen thuộc của BĐT Cô-si:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow (a+b+c)^2\geq 3(ab+bc+ac)\)
\(\Rightarrow 2(a+b+c)^2\geq 6(ab+bc+ac)(**)\)
Từ \((*); (**)\Rightarrow B\geq \frac{9}{(a+b+c)^2+2(a+b+c)^2}=\frac{3}{(a+b+c)^2}\geq \frac{3}{3^2}=\frac{1}{3}\)
(do \(a+b+c\leq 3)\)
Do đó: \(B_{\min}=\frac{1}{3}\)
Dấu bằng xảy ra khi \(a=b=c=1\)
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 \(\Leftrightarrow t^2=1\Leftrightarrow t=1\Leftrightarrow2a=b+c\Leftrightarrow a=b=c\)