chứng minh rằng:
\(\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{a+3c}\ge\dfrac{1}{a+2c+b}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\)
Giúp mình với, mình đang cần gấp...:))
Cho a,b,c là các số thực dương. Chứng minh rằng:
\(\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\le\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\)
Cho a,b,c>0. CM: \(\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\ge\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\)
Áp dụng bất đẳng thức \(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) với x, y, z > 0 ta có:
\(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}=\dfrac{1}{9}\left(\dfrac{9}{a+a+b}+\dfrac{9}{b+b+c}+\dfrac{1}{c+c+a}\right)\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{1}{9}.3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\).
cho a,b,c là các số thực dương thỏa mãn \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=2017\)
Tìm max \(P=\dfrac{1}{2a+3b+3c}+\dfrac{1}{3a+2b+3c}+\dfrac{1}{3a+3b+2c}\)
cho a,b,c là các số dương thay đổi thỏa mãn:
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=2017\)
Tìm GTLN của P biết : \(P=\dfrac{1}{2a+3b+3c}+\dfrac{1}{3a+2b+3c}+\dfrac{1}{3a+3b+2c}\)
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{b+c}\ge\dfrac{16}{2a+3b+3c}\)
\(\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{a+c}\ge\dfrac{16}{2b+3a+3c}\)
\(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{a+b}\ge\dfrac{16}{2c+3a+3b}\)
cộng tất cả lại ta được \(4.2017\ge16.\left(\dfrac{1}{2a+3b+3c}+\dfrac{1}{2b+3a+3c}+\dfrac{1}{2c+3a+3b}\right)< =>P\le\dfrac{2017}{4}\)
dấu bằng xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{a+b}=\dfrac{1}{b+c}=\dfrac{1}{a+c}\\\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=2017\end{matrix}\right.< =>\left\{{}\begin{matrix}a=b=c\\\dfrac{3}{2a}=\dfrac{3}{2b}=\dfrac{3}{2c}=2017\end{matrix}\right.< =>a=b=c=\dfrac{3}{4034}}\)
bài 1 : cho a, b, c>0 thỏa mãn a2+b2+c2=3
chứng minh rằng \(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}>=\dfrac{3}{2}\)
bài 2 : cho a, b, c>0. chứng minh rằng
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}>=\dfrac{1}{2}\)
bài 3 : cho a, b, c>0 thỏa mãn ab+bc+ac=abc
tìm GTLN của \(S=\dfrac{1}{3a+2b+c}+\dfrac{1}{3b+2c+a}+\dfrac{1}{3c+2a+b}\)
Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
1. Cho \(a,b,c>0\) và \(ab+bc+ca=abc\). Chứng minh rằng:
\(\dfrac{1}{a+3b+2c}+\dfrac{1}{b+3c+2a}+\dfrac{1}{c+3a+2b}\le\dfrac{1}{6}\)
2. Cho \(a,b\ge0\) và \(a+b=2\) Tìm Max
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+20ab\)
Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)
CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)
\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)
Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu = xảy ra khi a=b=c=3
Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)
\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)
Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)
\(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)
\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)
Vậy...
2,
\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)
\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(ab=x\Rightarrow0\le x\le1\)
\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)
\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)
Cho a,b,c>0. CMR
\(\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\le\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\)
\(\dfrac{1}{a+3b}+\dfrac{1}{a+b+2c}\ge\dfrac{4}{2a+4b+2c}=\dfrac{2}{a+2b+c}\)
Tương tự: \(\dfrac{1}{b+3c}+\dfrac{1}{b+c+2a}\ge\dfrac{2}{a+b+2c}\)
\(\dfrac{1}{c+3a}+\dfrac{1}{a+c+2b}\ge\dfrac{2}{2a+b+c}\)
Cộng vế với vế và rút gọn:
\(\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\)
Dấu "=" xảy ra khi \(a=b=c\)
Cho a,b,c>0. Chứng minh \(\dfrac{ab}{a+3b+2c}\)+\(\dfrac{bc}{b+3c+2a}\)+\(\dfrac{ca}{c+3c+2b}\)≤\(\dfrac{a+b+c}{6}\)
Mong mọi người giúp đỡ
Áp dụng BĐT
\(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\\ \Rightarrow\dfrac{9abc}{a+3a+2c}\\ =\dfrac{9}{\left(a+c\right)\left(b+c\right)+2b}\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{4}{2}\)
Tương tự với 2 BĐT còn lại rồi cộng vế theo vế
=> 9 vế trái
\(\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\\ +\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{a+b+c}{2}\\ =\dfrac{3\left(a+b+c\right)}{2}\\ \Rightarrow......._{\left(đpcm\right)}\)
Cho các số thực dương \(a;b;c\) thỏa mãn :\(ab+bc+ca=abc\). Chứng minh rằng :
\(\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{1}{6}\).
P/s: Em xin phép nhờ quý thầy cô và các bạn bè hỗ trợ và giúp đỡ với ạ. Em cám ơn rất nhiều!
\(ab+bc+ca=abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Đặt vế trái của BĐT cần chứng minh là P
Ta có:
\(\dfrac{1}{a+2b+3c}=\dfrac{1}{a+b+b+c+c+c}\le\dfrac{1}{6^2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+2b+3c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}\right)\)
Tương tự:
\(\dfrac{1}{b+2c+3a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{3}{a}\right)\) ; \(\dfrac{1}{c+2a+3b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{2}{a}+\dfrac{3}{b}\right)\)
Cộng vế:
\(P\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)