Cho a,b,c,d >0 .Chứng minh rằng...
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Cho a,b,c,d >0 .Chứng minh rằng : ab+2c+3d+bc+2d+3a+cd+2a+3b+da+2b+3c≥23
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BM
0
BM
0
27 tháng 8 2023
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a+2b}{a}=\dfrac{3bk+2b}{bk}=\dfrac{3k+2}{k}\)
\(\dfrac{3c+2d}{c}=\dfrac{3dk+2d}{dk}=\dfrac{3k+2}{k}\)
Do đó: \(\dfrac{3a+2b}{a}=\dfrac{3c+2d}{c}\)
b: \(\dfrac{2a-3b}{b}=\dfrac{2bk-3b}{b}=2k-3\)
\(\dfrac{2c-3d}{d}=\dfrac{2dk-3d}{d}=2k-3\)
Do đó: \(\dfrac{2a-3b}{b}=\dfrac{2c-3d}{d}\)
c: \(\dfrac{a}{a-2b}=\dfrac{bk}{bk-2b}=\dfrac{k}{k-2}\)
\(\dfrac{c}{c-2d}=\dfrac{dk}{dk-2d}=\dfrac{k}{k-2}\)
Do đó: \(\dfrac{a}{a-2b}=\dfrac{c}{c-2d}\)
BG
1
1 tháng 2 2017
a )\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{2a}{2c}\)
\(\frac{a-b}{c-d}=\frac{2a}{2c}\Rightarrow\frac{a-b}{2a}=\frac{c-d}{2c}\) ( đpcm)
b ) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{3a}{3c}=\frac{2b}{2d}=\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\)
\(\Rightarrow\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\) ( đpcm )
25 tháng 9 2017
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\)
\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)
\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)
*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)
4 tháng 7 2021
\(S=\left(1+\dfrac{2a}{3b}\right)\left(1+\dfrac{2b}{3c}\right)\left(1+\dfrac{2c}{3d}\right)\left(1+\dfrac{2d}{3a}\right)\)
có \(1+\dfrac{2a}{3b}\ge2\sqrt{\dfrac{2a}{3b}}\)(BDT AM-GM)
\(=>1+\dfrac{2b}{3c}\ge2\sqrt{\dfrac{2b}{3c}}\)
\(=>1+\dfrac{2c}{3d}\ge2\sqrt{\dfrac{2c}{3d}}\)
\(=>1+\dfrac{2d}{3a}\ge2\sqrt{\dfrac{2d}{3a}}\)
\(=>S\ge16\sqrt{\dfrac{2a.2b.2c.2d}{3a.3b.3c.3d}}=16\sqrt{\dfrac{16abcd}{81abcd}}=16\sqrt{\dfrac{16}{81}}=\dfrac{64}{9}\)
11 tháng 11 2023
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)