Cho a>=b>=c>0. Chứng minh rằng: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
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29 tháng 2 2020
Bài này anh Alibaba có trả lời bên h rồi,mik viết lại bạn dễ coi nha !
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\)
\(=\frac{a^2}{ab}+\frac{b^2}{bc}+\frac{c^2}{ca}+\frac{b^2}{b^2}\)
\(\ge\frac{\left(a+2b+c\right)^2}{ab+bc+ca+b^2}\)
\(=\frac{\left(a+b\right)^2+2\left(a+b\right)\left(b+c\right)+\left(b+c\right)^2}{\left(a+b\right)\left(b+c\right)}\)
\(=\frac{a+b}{b+c}+\frac{b+c}{a+b}+2\)
Anh ấy bảo đến đây bí và mik cũng như vậy T_T
KK
30 tháng 1 2017
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{a+c}+b+\frac{c^2}{a+b}+c\ge\frac{a+b+c}{2}+a+b+c\)
\(\Leftrightarrow a\left(\frac{a}{b+c}+1\right)+b\left(\frac{b}{a+c}+1\right)+c\left(\frac{c}{a+b}+1\right)\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow a\left(\frac{a+b+c}{b+c}\right)+b\left(\frac{a+b+c}{c+a}\right)+c\left(\frac{a+b+c}{a+b}\right)\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)\frac{a}{b+c}+\left(a+b+c\right)\frac{b}{c+a}+\left(a+b+c\right)\frac{c}{a+b}\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\ge\frac{3}{2}+3\)
\(\Leftrightarrow\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\ge\frac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge\frac{9}{2}\)
\(\Leftrightarrow2\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9\)
\(\Leftrightarrow\left(2a+2b+2c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\)
\(\Leftrightarrow\left(b+c+c+a+a+b\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}b+c+c+a+a+b\ge3\sqrt[3]{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\\\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\ge3\sqrt[3]{\frac{1}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}}\end{matrix}\right.\)
Nhân từng vế :
\(\Rightarrow\left(b+c+c+a+a+b\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\sqrt[3]{\left(b+c\right)\left(c+a\right)\left(a+b\right).\frac{1}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}}\)
\(\Rightarrow\left(b+c+c+a+a+b\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\left(đpcm\right)\)
Vậy với a ,b ,c > 0 thì \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
BN
30 tháng 1 2017
Áp dụng bất đẳng thức cô-si cho các số thực không âm ta có:
\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}\times\frac{b+c}{4}}=a\) (1)
\(\frac{b^2}{a+c}+\frac{a+c}{4}\ge2\sqrt{\frac{b^2}{a+c}\times\frac{a+c}{4}}=b\) (2)
\(\frac{c^2}{a+b}+\frac{a+b}{4}\ge2\sqrt{\frac{c^2}{a+b}\times\frac{a+b}{4}}=c\) (3)
Cộng (1),(2) và (3),vế theo vế ta được:
\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+\frac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\) (đpcm)
Dấu "=" xảy ra khi :a=b=c
Vậy \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\) với a,b,c >0
CV
15 tháng 12 2016
mình nghĩ đề bài sai một chỗ :\(\frac{a^2}{b^2}\)chứ ko phải là \(\frac{a}{b^2}\)
21 tháng 4 2018
\(\frac{a}{b^2}+\frac{1}{a}\ge\frac{2}{b}\) BĐT Cô-si
Tương tự suy ra đpcm
22 tháng 3 2019
Áp dụng bất đẳng thức Cô-si ta có:
\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
Cộng theo vế ta được:
\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 5 2021
Ta chứng minh BĐT sau với các số dương:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Áp dụng:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế với vế:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 5 2021
b.
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)
Cộng vế với vế (1); (2) và (3):
\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
xem lại đề đi bạn sai dấu thì phải
Xét hiệu:
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-\frac{c}{b}-\frac{a}{c}=\frac{a-c}{b}+\frac{b-a}{c}+\frac{c-b}{a}\)
\(=\frac{ca.\left(a-c\right)}{abc}+\frac{ab.\left(b-a\right)}{abc}+\frac{bc.\left(c-b\right)}{abc}\)\(=\frac{a^2c-c^2a}{abc}+\frac{b^2a-a^2b}{abc}+\frac{c^2b-b^2c}{abc}\)
\(=\frac{a^2c-c^2a+b^2a-a^2b+c^2b-b^2c}{abc}\)\(=\frac{\left(a^2c-b^2c\right)+\left(-c^2a+c^2b\right)+\left(b^2a-a^2b\right)}{abc}\)
\(=\frac{c.\left(a-b\right)\left(a+b\right)-c^2.\left(a-b\right)-ab.\left(a-b\right)}{abc}\)\(=\frac{\left(a-b\right)\left[c.\left(a+b\right)-c^2-ab\right]}{abc}\)
\(=\frac{\left(a-b\right)\left(ac+bc-c^2-ab\right)}{abc}\)\(=\frac{\left(a-b\right)\left[\left(ac-c^2\right)+\left(bc-ab\right)\right]}{abc}\)
\(=\frac{\left(a-b\right)\left[c.\left(a-c\right)-b.\left(a-c\right)\right]}{abc}\)\(=\frac{\left(a-b\right)\left(a-c\right)\left(c-b\right)}{abc}\)
ta thấy \(a\ge b\ge c>0\Rightarrow abc>0\)
\(a-b\ge0\left(a\ge b\right);a-c\ge0\left(a\ge b\ge c\right);c-b\le0\left(b\ge c\right)\)\(\Rightarrow\left(a-b\right)\left(a-c\right)\left(c-b\right)\le0\)
\(\text{Suy ra: }\frac{\left(a-b\right)\left(a-c\right)\left(c-b\right)}{abc}\le0\)
\(\Rightarrow\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\le\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
có thể sai đề