1. Cho a,b,c > 0. Cmr :
\(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
2. Cho a,b,c > 0. Cmr :
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge\frac{2}{3}\)
1.
\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)
\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Dấu "=" khi \(a=b=c\)
2.
\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" khi \(a=b=c=d\)
Thục Trinh, tran nguyen bao quan, Phùng Tuệ Minh, Ribi Nkok Ngok, Lê Nguyễn Ngọc Nhi, Tạ Thị Diễm Quỳnh,
Nguyễn Huy Thắng, ?Amanda?, saint suppapong udomkaewkanjana
Help me!
Bài thứ hai đó áp dụng bđt cauchy showas là ra rồi sử dụng tch bắc cầu tệ.
Cho \(\frac{a}{b}\)=\(\frac{c}{d}\)
a)C/m \(\frac{a-b}{b}\)=\(\frac{c-d}{d}\)
b)C/m \(\frac{a+2c}{b-2d}\)=\(\frac{3a-4c}{3b-4d}\)
c)\(\frac{\left(a+b\right)^3}{a^3+b^3}\)=\(\frac{\left(c+d\right)^3}{c^3+d^3}\)
cho a,b,c,d > 0. CMR \(\frac{a^4}{a^3+2b^3}+\frac{b^4}{b^3+2c^3}+\frac{c^4}{c^3+2d^3}+\frac{d^4}{d^3+2a^3}\ge\frac{a+b+c+d}{3}\)
Cho a, b, c, d là các số thực dương. Chứng minh :
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge\frac{2}{3}\)
Cho a, b, c, d > 0. CMR: \(\frac{a^4}{a^3+2b^3}+\frac{b^4}{b^3+2c^3}+\frac{c^4}{c^3+2d^3}+\frac{d^4}{d^3+2a^3}\ge\frac{a+b+c+d}{3}\) (Dùng Cô-si )
Bạn tham khảo (hoàn toàn dùng Cô-si):
Câu hỏi của Trần Anh Thơ - Toán lớp 8 | Học trực tuyến
bài 1: cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)
a) CMR: (a+2c)(b+d)=(a+c)(b+2d) \(\left(b,d\ne0\right)\)
b) CMR: (a+c)(b-d)=ab-cd
c) CMR: \(\frac{a}{a-b}=\frac{c}{c-d}\left(a,b,c,d>0;a\ne b,c\ne d\right)\)
bài 2: cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}CMR:\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
cho a,b,c,d >0 . cmr:
\(\frac{a}{b+2c+3d}\) +\(\frac{b}{c+2d+3a}\)+\(\frac{c}{d+2a+3b}\)+\(\frac{d}{a+2b+3c}\)\(\ge\) \(\frac{2}{3}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\). Mà theo BĐT AM-GM ta có:
\(\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}=\frac{\left(a+b+c+d\right)^2}{2\left[\left(a+b\right)\left(c+d\right)+\left(a+c\right)\left(b+d\right)+\left(a+d\right)\left(b+c\right)\right]}\ge\frac{2}{3}\)
Đẳng thức xảy ra khi a=b=c=d
cho \(\frac{a}{b}\)=\(\frac{c}{d}\) CMR
A)\(\frac{3a-2c}{3b+2d}\)= \(\frac{2a+5c}{2b+5d}\)
B)\(\frac{a^3}{b^3}\) =\(\frac{\left(a+c\right)^3}{\left(b+d\right)^5}\)
Đặt
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
a)
Sửa đề nhá :
\(\frac{3a+2c}{3b+2d}=\frac{3bk+2dk}{3b+2d}=\frac{k\left(3b+2d\right)}{\left(3b+2d\right)}=k\)(1)
\(\frac{2a+5c}{2b+5d}=\frac{2bk+5dk}{2b+5d}=\frac{k\left(2b+5d\right)}{2b+5d}=k\)(2)
Từ (1) và (2)
=> \(\frac{3a+2c}{3b+2d}=\frac{2a+5c}{2b+5d}\)
b)
\(\frac{a^3}{b^3}=\frac{b^3k^3}{b^3}=k^3\)(3)
\(\frac{\left(a+c\right)^3}{\left(b+d\right)^3}=\frac{\left(bk+dk\right)^3}{\left(b+d\right)^3}=\frac{k^3\left(b+d\right)^3}{\left(b+d\right)^3}=k^3\)(4)
Từ (3) và (4)
=> \(\frac{a^3}{b^3}=\frac{\left(a+c\right)^3}{\left(b+d\right)^3}\)
Cho \(a;b;c;d>0\)chứng minh
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge\frac{2}{3}\)
cre: dự vào đề tóan quốc tế mỹ
\(\text{Σ}\frac{a}{b+2c+3d}=\text{Σ}\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{6\left(ab+bc+cd+ad\right)}\)
\(=\frac{\left(a+b\right)^2+\left(c+d\right)^2+2\left(a+b\right)\left(c+d\right)}{6\left(ab+bc+cd+ad\right)}=\frac{a^2+c^2+b^2+d^2+2ab+2cd+2\left(a+b\right)\left(c+d\right)}{6\left(ab+bc+cd+ad\right)}\)
\(\ge\frac{4\left(ab+bc+cd+ad\right)}{6\left(ab+bc+cd+ad\right)}=\frac{2}{3}\)
Dấu = xảy ra khi a=b=c=d
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\)
\(=\frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\frac{d^2}{ad+2bd+3cd}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{4.\left(ab+ad+bc+bd+ca+cd\right)}\)\(\ge\frac{\left(a+b+c+d\right)^2}{\frac{3}{2}.\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=d\)
\(VT=\frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\frac{d^2}{ad+2bd+3cd}\)
Áp dụng BĐT Svac-xơ cho 3 số dương ta được :
\(VT\ge\frac{\left(a+b+c+d\right)^2}{4ab+4ac+4ad+4bc+4bd+4cd}\)
Áp dụng BĐT phụ \(x^2+y^2\ge2xy\) ta được :
\(a^2+b^2\ge2ab;a^2+c^2\ge2ac;a^2+d^2\ge2ad\)
\(b^2+c^2\ge2bc;b^2+d^2\ge2bd;c^2+d^2\ge2cd\)
\(\Rightarrow3\left(a^2+b^2+c^2+d^2\right)\ge2\left(ab+ac+ad+bc+bd+cd\right)\)
Ta lại có : \(\left(a+b+c+d\right)^4=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd\)
\(\ge\frac{8\left(ab+ac+ad+bc+bd+cd\right)}{3}\)
\(\Rightarrow VT\ge\frac{\left(a+b+c+d\right)^4}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{8\left(ab+ac+ad+bc+bd+cd\right)}{12\left(ab+ac+ad+bc+bd+cd\right)}=\frac{2}{3}\)
Dấu "=" xảy ra khi \(a=b=c=d\)
