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đặt 6a=x;2b=y;3c=z=>x+y+z=11
áp dụng bất đẳng thức Schwarts ta có:\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+y+z+3}=\frac{9}{14}\)
\(\Leftrightarrow\frac{28}{x+1}+\frac{28}{y+1}+\frac{28}{z+1}\ge\frac{28.9}{14}=18\)
\(\Leftrightarrow\frac{28}{x+1}-1+\frac{28}{y+1}-1+\frac{28}{z+1}-1\ge18-1-1-1=15\)
\(\Leftrightarrow\frac{27-x}{x+1}+\frac{27-y}{y+1}+\frac{27-z}{z+1}\ge15\)
\(\Leftrightarrow\frac{11-x+16}{x+1}+\frac{11-y+16}{y+1}+\frac{11-z+16}{z+1}\ge15\)
\(\Leftrightarrow\frac{y+z+16}{x+1}+\frac{z+x+16}{y+1}+\frac{x+y+16}{z+1}\ge15\)
\(\Leftrightarrow\frac{2b+3c+16}{6a+1}+\frac{6a+3c+16}{2b+1}+\frac{6a+2b+16}{3c+1}\ge15\)
=>đpcm
dấu "=" xảy ra khi \(a=\frac{11}{18};b=\frac{11}{6};c=\frac{11}{9}\)
Theo BĐT Bu nhi a cốp xki ta có :
\(\left(a+b+c+d\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\ge16\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{16}{a+b+c+d}\)
Áp dụng vào bài toán ta có :
\(\dfrac{1}{3a+3b+2c}=\dfrac{1}{16}.\dfrac{16}{\left(a+b\right)+\left(a+b\right)+\left(b+c\right)+\left(c+a\right)}\le\dfrac{1}{16}\left(\dfrac{1}{a+b}+\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
\(\dfrac{1}{3b+3c+2a}=\dfrac{1}{16}.\dfrac{16}{\left(b+c\right)+\left(b+c\right)+\left(a+b\right)+\left(c+a\right)}\le\dfrac{1}{16}\left(\dfrac{1}{b+c}+\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{c+a}\right)\)
\(\dfrac{1}{3c+3a+2b}=\dfrac{1}{16}.\dfrac{16}{\left(c+a\right)+\left(c+a\right)+\left(a+b\right)+\left(b+c\right)}\le\dfrac{1}{16}\left(\dfrac{1}{c+a}+\dfrac{1}{c+a}+\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\)
Cộng từng vế của BĐT ta được :
\(\dfrac{1}{3a+3b+2c}+\dfrac{1}{3b+3c+2a}+\dfrac{1}{3c+3a+2b}\le\dfrac{1}{16}\left(\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\right)=\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{1}{4}.6=\dfrac{3}{2}\)
Vậy GTLN của A là \(\dfrac{3}{2}\) . Dấu \("="\) xảy ra khi \(a=b=c=\dfrac{1}{4}\)
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
\(\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\\ +\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{a+b+c}{2}\\ =\dfrac{3\left(a+b+c\right)}{2}\\ \Rightarrow......._{\left(đpcm\right)}\)
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:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
á mk xl nhá mk ko đọc kĩ đề mk làm nhầm rùi bài mk làm là tìm GTNN nhá bạn ( mất công quá
)
ta có A= a+b+c+\(\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
= \(\dfrac{3a}{4}+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c}{4}+\dfrac{3c}{4}+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
=\(\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}\)
vì a,b,c >0 ===> \(\dfrac{3a}{4}>0,\dfrac{3}{a}>0,\dfrac{b}{2}>0,\dfrac{9}{2b}>0,\dfrac{c}{4}>0,\dfrac{4}{c}>0\)
áp dụng BĐT côsi cho các cặp số dương ta đc:
\(\dfrac{3a}{4}+\dfrac{3}{a}>=2.\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}=3\)
\(\dfrac{b}{2}+\dfrac{9}{2b}>=3\)(làm như trên nhá)
\(\dfrac{c}{4}+\dfrac{4}{c}>=2\)
===> \(\dfrac{3a}{4}+\dfrac{3}{a}+\dfrac{b}{2}+\dfrac{9}{2b}+\dfrac{c}{4}+\dfrac{4}{c}>=8\left(1\right)\)
có: \(\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}=\dfrac{a+2b+3c}{4}\)
mà a+2b+3c >= 20
===> \(\dfrac{a+2b+3c}{4}>=\dfrac{20}{4}=5\)
===> \(\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}>=5\left(2\right)\)
từ (1) và(2)===> a+b+c+\(\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}>=13\)
===> A >= 13
Dấu ''='' xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{3a}{4}=\dfrac{3}{a}\\\dfrac{b}{2}=\dfrac{9}{2b}\\\dfrac{c}{4}=\dfrac{4}{c}\\a+2b+3c=20\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
Vậy Min A=13 <=>\(\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
Ối,không ngờ đề gắt ~v
Theo Cô si,ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{3}{\sqrt[3]{xyz}}\ge\frac{3}{\frac{x+y+z}{3}}=\frac{9}{x+y+z}\)
Suy ra \(\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Áp dụng vào,ta có: \(\frac{1}{a+2b+3c}=\frac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\)
\(\le\frac{1}{9}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{b+c}\right)\)
Chứng minh tương tự và cộng theo vế:
\(VT\le\frac{1}{9}\left[\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)+2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\right]\)
\(=\frac{1}{9}\left[3\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\right]=\frac{1}{3}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
Lại có BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Áp dụng vào,ta có: \(VT\le\frac{1}{3}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(\le\frac{1}{12}\left[2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Nhân abc vào mỗi vế : \(VT.abc\le\frac{1}{6}\left(ab+bc+ca\right)=\frac{abc}{6}\)
Chia cả hai vế cho abc (vì a,b,c dương nên abc khác 0): \(VT\le\frac{1}{6}< \frac{3}{16}\)(đpcm)
Cũng không biết đúng hay sai nữa :v