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Giải:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\) \(\begin{cases}a=bk\\c=dk\end{cases}\)
Thay vào vế trái ta có:
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)
Thay vào vế phải ta có:
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)
\(\Rightarrow VP=VT=\dfrac{2k+3}{2k-3}\Rightarrow\) Đpcm
Ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{3b}{3d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{3b}{3d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\)
\(\Rightarrow\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\Rightarrow\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\) (ĐPCM)
Ta có:
a/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a}{3b}=\dfrac{2c}{2d}=\dfrac{3a+2c}{3b+2d}\)
b/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{-2a}{-2b}=\dfrac{7c}{7d}=\dfrac{-2a+7c}{-2b+7d}\)
PS: Xong
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 \(a=b=c=\dfrac{1}{4}\)
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:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
\(\text{Ta có : }\dfrac{2a+b}{a+b}+\dfrac{2b+c}{b+c}+\dfrac{2c+d}{c+d}+\dfrac{2d+a}{d+a}=6\\ \Rightarrow\left[\left(\dfrac{2a+b}{a+b}-1\right)+\left(\dfrac{2b+c}{b+c}-1\right)-1\right]+\left[\left(\dfrac{2c+d}{c+d}-1\right)+\left(\dfrac{2d+a}{d+a}-1\right)-1\right]=0\\ \Rightarrow\left(\dfrac{a}{a+b}+\dfrac{b}{b+c}-1\right)+\left(\dfrac{c}{c+d}+\dfrac{d}{d+a}-1\right)=0\\ \Rightarrow\left(\dfrac{a\left(b+c\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{b\left(a+b\right)}{\left(a+b\right)\left(b+c\right)}-\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)}\right)+\left(\dfrac{c\left(d+a\right)}{\left(c+d\right)\left(d+a\right)}+\dfrac{d\left(c+d\right)}{\left(c+d\right)\left(d+a\right)}-\dfrac{\left(c+d\right)\left(d+a\right)}{\left(c+d\right)\left(d+a\right)}\right)=0\\ \Rightarrow\dfrac{ab+ac+ab+b^2-ab-b^2-ac-bc}{\left(a+b\right)\left(b+c\right)}+\dfrac{cd+ac+cd+d^2-cd-d^2-ac-ad}{\left(c+d\right)\left(d+a\right)}=0\\ \Rightarrow\dfrac{ab-bc}{\left(a+b\right)\left(b+c\right)}+\dfrac{cd-ad}{\left(c+d\right)\left(d+a\right)}=0\)\(\Rightarrow\dfrac{ab-bc}{\left(a+b\right)\left(b+c\right)}=\dfrac{ad-cd}{\left(c+d\right)\left(d+a\right)}\\ \Rightarrow\dfrac{b\left(a-c\right)}{\left(a+b\right)\left(b+c\right)}=\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}\\ \Rightarrow\dfrac{b}{\left(a+b\right)\left(b+c\right)}=\dfrac{d}{\left(c+d\right)\left(d+a\right)}\left(Vìa;b;c;d>0\right)\\ \Rightarrow b\left(c+d\right)\left(d+a\right)=d\left(a+b\right)\left(b+c\right)\\ \Rightarrow\left(bc+bd\right)\left(d+a\right)=\left(ad+bd\right)\left(b+c\right)\)
\(\Rightarrow bcd+bd^2+abc+abd=abd+b^2d+acd+bcd\\ \Rightarrow bd^2-b^2d=acd-abc\\ \Rightarrow bd\left(d-b\right)=ac\left(d-b\right)\\ \Rightarrow bd=ac\left(Vìd-b\ne0\right)\\ \Rightarrow abcd=ac\cdot bd=ac\cdot ac=\left(ac\right)^2\)
Vậy \(abcd\) là số chính phương
Bài 1:
Áp dụng BĐt cauchy dạng phân thức:
\(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\ge\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3x+3y}=4\)
dấu = xảy ra khi 2x+y=x+2y <=> x=y
Bài 2:
ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{4^2}{a+b+c+d}=\dfrac{16}{a+b+c+d}\)(theo BĐt cauchy-schwarz)
\(\Rightarrow\dfrac{1}{a+b+c+d}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\)
Áp dụng BĐT trên vào bài toán ta có:
\(A=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)\(A\le\dfrac{1}{16}.4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
......
dấu = xảy ra khi a=b=c
Bài 2:
Áp dụng BĐT cauchy cho 2 số dương:
\(a^2+1\ge2a\)
\(\Leftrightarrow\dfrac{a}{a^2+1}\le\dfrac{a}{2a}=\dfrac{1}{2}\)
thiết lập tương tự:\(\dfrac{b}{b^2+1}\le\dfrac{1}{2};\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)
cả 2 vế các BĐT đều dương ,cộng vế với vế,ta có dpcm
dấu = xảy ra khi a=b=c=1
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 <=>\(\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
Bài 1:
a). Ta có: a < b
=> -6a > -6b
mà 3 > 1
=> \(3-6a>1-6b\)
b)
Ta có: a < b
=> a - 2 < b - 2
=> \(7\left(a-2\right)< 7\left(b-2\right)\)
c)
Ta có: a < b
=> -2a > -2b
=> 1 - 2a > 1 - 2b
\(\Rightarrow\dfrac{1-2a}{3}>\dfrac{1-2b}{3}\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c+2a-3c}{2b+3d+2b-3d}=\dfrac{a}{b}\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c-\left(2a-3c\right)}{2b+3d-\left(2b-3d\right)}=\dfrac{c}{d}\)
Suy ra \(\dfrac{a}{b}=\dfrac{c}{d}\)