Cho các số thực thỏa mãn\(\dfrac{â}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}chứngminh:\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)
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thì nó khó thiệt mà
9 tháng 12 2021
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{a+b+c+d}=1\\ \Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Rightarrow a=b=c=d\\ \Rightarrow VT=\left(\dfrac{2019a+2020a-2021a}{2019a+2020a-2021a}\right)^3=1^3=1=\dfrac{a^2}{a\cdot a}=VP\)
10 tháng 12 2021
Sửa đề: \(\dfrac{2018a-2019b}{2019a+2020b}=\dfrac{2018c-2019d}{2019c+2020d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2020a}{2020b}=\dfrac{2020c}{2020d}=\dfrac{2019a}{2019c}=\dfrac{2019b}{2019d}=\dfrac{2018a}{2018c}=\dfrac{2018b}{2018d}=\dfrac{2018a-2019b}{2018c-2019d}=\dfrac{2019a+2020b}{2019c+2020d}\\ \Leftrightarrow\dfrac{2018a-2019b}{2019a+2020b}=\dfrac{2018c-2019d}{2019c+2020d}\)
8 tháng 1 2021
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Leftrightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)
\(\Leftrightarrow\dfrac{2020a^2}{2020c^2}=\dfrac{2021b^2}{2021d^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2020a^2}{2020c^2}=\dfrac{2021b^2}{2021d^2}=\dfrac{2020a^2+2021b^2}{2020c^2+2021d^2}=\dfrac{2020a^2-2021b^2}{2020c^2-2021d^2}\)
Ta có: \(\dfrac{2020a^2+2021b^2}{2020c^2+2021d^2}=\dfrac{2020a^2-2021b^2}{2020c^2-2021d^2}\)(cmt)
nên \(\dfrac{2020a^2+2021b^2}{2020a^2-2021b^2}=\dfrac{2020c^2+2021d^2}{2020c^2-2021d^2}\)(đpcm)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1 2021
Dự đoán điểm rơi xảy ra tại \(\left(a;b;c\right)=\left(3;2;4\right)\)
Đơn giản là kiên nhẫn tính toán và tách biểu thức:
\(D=13\left(\dfrac{a}{18}+\dfrac{c}{24}\right)+13\left(\dfrac{b}{24}+\dfrac{c}{48}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{2}{ab}\right)+\left(\dfrac{a}{18}+\dfrac{c}{24}+\dfrac{2}{ac}\right)+\left(\dfrac{b}{8}+\dfrac{c}{16}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{c}{12}+\dfrac{8}{abc}\right)\)
Sau đó Cô-si cho từng ngoặc là được
27 tháng 11 2021
1.
Đặt \(\left(x;y;z\right)=\left(\dfrac{a}{a+b};\dfrac{b}{b+c};\dfrac{c}{c+a}\right)\Rightarrow\left\{{}\begin{matrix}1-x=\dfrac{b}{b+a}\\1-y=\dfrac{c}{b+c}\\1-z=\dfrac{a}{a+c}\end{matrix}\right.\)
\(\Rightarrow xyz=\dfrac{1}{8}\\ xyz=\left(1-x\right)\left(1-y\right)\left(1-z\right)\\ \Rightarrow xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)-xyz\\ \Rightarrow2xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)=\dfrac{1}{4}\\ \Rightarrow x+y+z=\dfrac{3}{4}+xy+yz+zx\)
\(\RightarrowĐpcm\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2021
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2021
\(\Leftrightarrow\left(1+ab+bc+ca\right)\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(a+b\right)\left(b+c\right)}+\dfrac{1}{\left(a+c\right)\left(b+c\right)}\right)\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
Áp dụng BĐT quen thuộc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(ab+bc+ca\right)\left(a+b+c\right)=\dfrac{8}{9}\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\)
Ta chỉ cần chứng minh:
\(\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow4\left(ab+bc+ca\right)^2\ge9abc+9abc\left(ab+bc+ca\right)\)
Do \(3\left(ab+bc+ca\right)^2\ge9abc\left(a+b+c\right)=9abc\)
Nên ta chỉ cần chứng minh:
\(\left(ab+bc+ca\right)^2\ge9abc\left(ab+bc+ca\right)\)
\(\Leftrightarrow ab+bc+ca\ge9abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
Hiển nhiên đúng do \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\left(1\right)\\ \text{Đặt }\dfrac{a}{b}=\dfrac{b}{c}=k\Leftrightarrow a=bk;b=ck\Leftrightarrow a=ck^2\\ \Leftrightarrow\dfrac{a^2}{bc}=\dfrac{c^2k^4}{c^2k}=k^3=\left(\dfrac{a}{b}\right)^3\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)