cho a,b,c là các số thực dương thay đổi bất kì
cm:
\(\dfrac{\left(2a+b+c\right)^2}{2a^2+\left(b+c\right)^2}+\dfrac{\left(a+2b+c\right)^2}{2b^2+\left(c+a\right)^2}+\dfrac{\left(a+b+2c\right)^2}{2c^2+\left(a+b\right)^2}\le8\)
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NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 5 2021
Bài này có bạn giải rồi:
Cho các số thực dương a,b,c.Chứng minh rằng :\(\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}+\dfrac{c\left(2b-c\right)}{... - Hoc24
29 tháng 3 2021
\(\Leftrightarrow\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}-2+\dfrac{c\left(2b-c\right)}{b\left(c+a\right)}-2+\dfrac{a\left(2c-a\right)}{c\left(a+b\right)}-2\le\dfrac{3}{2}-6\)
\(\Leftrightarrow\dfrac{b^2+2ac}{a\left(b+c\right)}+\dfrac{c^2+2ab}{b\left(c+a\right)}+\dfrac{a^2+2bc}{c\left(a+b\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\dfrac{b^2}{ab+ac}+\dfrac{c^2}{bc+ab}+\dfrac{a^2}{ac+bc}+\dfrac{2c^2}{bc+c^2}+\dfrac{2a^2}{ac+a^2}+\dfrac{2b^2}{ab+b^2}\ge\dfrac{9}{2}\)
Ta có:
\(VT\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}+\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ca}\)
\(\Leftrightarrow VT\ge\left(a+b+c\right)^2\left(\dfrac{1}{2\left(ab+bc+ca\right)}+\dfrac{1}{a^2+b^2+c^2+ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2+ab+bc+ca}\right)\)
\(\Leftrightarrow VT\ge\dfrac{9\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)+2\left(a^2+b^2+c^2+ab+bc+ca\right)}\)
\(\Leftrightarrow VT\ge\dfrac{9\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}=\dfrac{9}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 8 2021
\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
Tương tự và cộng lại:
\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
24 tháng 8 2021
Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)
\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)
Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z
\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)
Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))
\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))
\(\Rightarrow P\le\dfrac{3}{16}\)
\(ĐTXR\Leftrightarrow a=b=c=1\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 12 2020
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge3+\dfrac{2a^2+2b^2+2c^2-2\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge5-\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)
Do \(\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}=\dfrac{2a^2}{ab+ac}+\dfrac{2b^2}{bc+ab}+\dfrac{2c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)
Điều này hiển nhiên đúng do:
\(VT=\dfrac{2}{3}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}+\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)
\(VT\ge2\sqrt{\dfrac{12\left(a+b+c\right)^2\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)\left(a+b+c\right)^2}}+\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=5\)
Dấu "=" xảy ra khi \(a=b=c\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1
Trước hết theo BĐT Schur bậc 3 ta có:
\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc\ge2\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3abc\ge2\left(ab+bc+ca\right)\) (do \(a+b+c=3\)) (1)
Đặt vế trái BĐT cần chứng minh là P, ta có:
\(P=\dfrac{\left(a^2+abc\right)^2}{a^2b^2+2abc^2}+\dfrac{\left(b^2+abc\right)^2}{b^2c^2+2a^2bc}+\dfrac{\left(c^2+abc\right)^2}{a^2c^2+2ab^2c}\)
\(\Rightarrow P\ge\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{\left(ab+bc+ca\right)^2}\)
Áp dụng (1):
\(\Rightarrow P\ge\dfrac{\left[2\left(ab+bc+ca\right)\right]^2}{\left(ab+bc+ca\right)^2}=4\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
C
13 tháng 1
Anh giúp em câu này ạ, câu này hơi khó anh ạ, làm chắc cũng lâu, có gì anh để mai cũng được ạ!
https://hoc24.vn/cau-hoi/cho-hinh-chop-sabcd-co-day-la-hinh-binh-hanh-m-va-p-la-hai-diem-lan-luot-di-dong-tren-ad-va-sc-sao-cho-mamd-pspc-x-x0-mat-phang-a-di-qua-m-va-song-song-voi-sab-cat-hinh-chop-sabcd-t.8753881358034
16 tháng 2 2018
Bao nhiêu công gõ bài xong rồi đi chơi, chơi về định gửi bài, chơi về bật máy lên gửi thì lỗi, may vãi
16 tháng 2 2018
Ta có:
\(\dfrac{a^2}{\left(2a+b\right)\left(2a+c\right)}=\dfrac{a^2}{2a\left(a+b+c\right)+2a^2+bc}\)
\(\le\dfrac{1}{9}\left(\dfrac{a^2}{a\left(a+b+c\right)}+\dfrac{a^2}{a\left(a+b+c\right)}+\dfrac{a^2}{2a^2+bc}\right)\)
\(=\dfrac{1}{9}\left(\dfrac{2a}{a+b+c}+\dfrac{a^2}{2a^2+bc}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{1}{9}\left(\dfrac{2\left(a+b+c\right)}{a+b+c}+\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\right)\)
\(=\dfrac{1}{9}\left(2+\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\right)\)
Cần chứng minh \(\dfrac{1}{9}\left(2+\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\right)\le\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\le1\)
\(\Leftrightarrow\dfrac{bc}{bc+2a^2}+\dfrac{ca}{ca+2b^2}+\dfrac{ab}{ab+2c^2}\ge1\)
Cauchy-Schwarz: \(VT=\dfrac{bc}{bc+2a^2}+\dfrac{ca}{ca+2b^2}+\dfrac{ab}{ab+2c^2}\)
\(=\dfrac{b^2c^2}{b^2c^2+2a^2bc}+\dfrac{c^2a^2}{c^2a^2+2ab^2c}+\dfrac{a^2b^2}{a^2b^2+2abc^2}\)
\(\ge\dfrac{\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2}=1\) * Đúng*
Happy New Year (Lunar)
Chuẩn hóa \(a+b+c=3\)
\(\dfrac{\left(2a+b+c\right)^2}{2a^2+\left(b+c\right)^2}=\dfrac{\left(a+3\right)^2}{2a^2+\left(3-a\right)^2}=\dfrac{a^2+6a+9}{3\left(a^2-2a+3\right)}=\dfrac{1}{3}\left(1+\dfrac{8a+6}{\left(a-1\right)^2+2}\right)\le\dfrac{1}{3}\left(1+\dfrac{8a+6}{2}\right)\)
Tương tự và cộng lại:
\(VT\le\dfrac{1}{3}\left(3+\dfrac{8\left(a+b+c\right)+18}{2}\right)=8\) (đpcm)
Tuyệt :>