a,b,c\(\ge\)0
CM \(\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ca}+\dfrac{c^2}{2c^2+ab}\le1\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
12 tháng 1 2021
Đặt \(\left(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\right)=\left(x,y,z\right)\) với x, y, z > 0 thì ta có \(x+y+z=1\).
Đặt biểu thức ở VT là A. Ta có:
\(A=\sqrt{\dfrac{b^2+2a^2}{a^2b^2}}+\sqrt{\dfrac{c^2+2b^2}{b^2c^2}}+\sqrt{\dfrac{a^2+2c^2}{c^2a^2}}=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\).
Ta có bất đẳng thức \(\sqrt{a_1^2+a_2^2}+\sqrt{a_3^2+a_4^2}\ge\sqrt{\left(a_1+a_3\right)^2+\left(a_2+a_4\right)^2}\).
Đây là bđt Mincopxki cho hai bộ số thực và dễ dàng cm bằng biến đổi tương đương.
Do đó \(A\ge\sqrt{\left(x+y\right)^2+\left(\sqrt{2}y+\sqrt{2}z\right)^2}+\sqrt{z^2+2x^2}\ge\sqrt{\left(x+y+z\right)^2+\left(\sqrt{2}y+\sqrt{2}z+\sqrt{2}x\right)^2}=\sqrt{1+2}=\sqrt{3}=VP\).
Đẳng thức xảy ra khi a = b = c = 3.
Vậy...
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1 2021
Tương tự: \(GT\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
\(VT=\dfrac{\sqrt{a^2+a^2+b^2}}{ab}+\dfrac{\sqrt{b^2+b^2+c^2}}{bc}+\dfrac{\sqrt{c^2+a^2+a^2}}{ca}\)
\(VT\ge\dfrac{\sqrt{\dfrac{1}{3}\left(a+a+b\right)^2}}{ab}+\dfrac{\sqrt{\dfrac{1}{3}\left(b+b+c\right)^2}}{bc}+\dfrac{\sqrt{\dfrac{1}{3}\left(c+c+a\right)^2}}{ca}\)
\(VT\ge\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=3\)
30 tháng 8 2021
\(VT=\sqrt{\dfrac{a^2b^2}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{b^2c^2}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{a^2c^2}{b\left(a+b+c\right)+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{ac+ab+bc+c^2}}+\sqrt{\dfrac{b^2c^2}{a^2+ac+ab+bc}}+\sqrt{\dfrac{a^2c^2}{ab+bc+b^2+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{\left(c+a\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2c^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{bc}{a+b}+\dfrac{bc}{a+c}}{2}\\\sqrt{\dfrac{a^2c^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(b+c\right)\left(a+c\right)}}\le\dfrac{\dfrac{ab}{b+c}+\dfrac{ab}{a+c}}{2}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ca}{b+c}+\dfrac{ab}{b+c}\right)+\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)}{2}\\ \Rightarrow VT\le\dfrac{a+b+c}{2}=\dfrac{2}{2}=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
\(P=\dfrac{a^2}{ab+\dfrac{1}{b}}+\dfrac{b^2}{bc+\dfrac{1}{c}}+\dfrac{c^2}{ca+\dfrac{1}{a}}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}\)
\(P\ge\dfrac{3\left(ab+bc+ca\right)}{ab+bc+ca+\dfrac{ab+bc+ca}{abc}}=\dfrac{3}{1+\dfrac{1}{abc}}=\dfrac{3abc}{1+abc}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
NB
27 tháng 1 2021
Với a, b, c > 0 có:
\(P=\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\\ =\dfrac{a^2}{a\left(b+2c\right)}+\dfrac{b^2}{b\left(c+2a\right)}+\dfrac{c^2}{c\left(a+2b\right)}\)
\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\)
chọn \(\alpha=\dfrac{1}{abc}\Rightarrow dpcm\)
RD
9 tháng 12 2018
a)Bunhia:
\(\left(1+2\right)\left(b^2+2a^2\right)\ge\left(1.b+\sqrt{2}.\sqrt{2}a\right)^2=\left(b+2a\right)^2\)
b)\(ab+bc+ca=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bđt câu a
=>VT\(\ge\)\(\dfrac{b+2a}{\sqrt{3}ab}+\dfrac{c+2b}{\sqrt{3}bc}+\dfrac{a+2c}{\sqrt{3}ca}\)
\(\Leftrightarrow VT\ge\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{2}{a}=3=VP\)
Tự tìm dấu "="
9 tháng 12 2018
Nguyễn Việt LâmMashiro ShiinaBNguyễn Thanh HằngonkingCẩm MịcFa CTRẦN MINH HOÀNGhâu DehQuân Tạ MinhTrương Thị Hải Anh
4 tháng 12 2018
\(A=\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\)
\(\Leftrightarrow2A=\dfrac{2a^2}{2a^2+bc}+\dfrac{2b^2}{2b^2+ac}+\dfrac{2c^2}{2c^2+ab}\)
\(=1-\dfrac{bc}{2a^2+bc}+1-\dfrac{ac}{2b^2+ac}+1-\dfrac{ab}{2c^2+ab}\)
\(=3-\dfrac{bc}{2a^2+bc}-\dfrac{ac}{2b^2+ac}-\dfrac{ab}{2c^2+ab}\)
CM: \(P=\dfrac{bc}{2a^2+bc}+\dfrac{ac}{2b^2+ac}+\dfrac{ab}{2c^2+ab}\ge1\)
Thật vậy:
\(P\ge\dfrac{\left(ab+bc+ac\right)^2}{2a^2bc+b^2c^2+2b^2ac+a^2c^2+2c^2ab+a^2b^2}\)
\(=\dfrac{\left(ab+bc+ac\right)^2}{a^2bc+a^2bc+b^2c^2+b^2ac+b^2ac+a^2c^2+c^2ab+c^2ab+a^2b^2}\)
\(=\dfrac{\left(ab+bc+ac\right)^2}{ab\left(ac+bc+ab\right)+bc\left(ab+bc+ac\right)+ac\left(ab+bc+ac\right)}\)
\(=1\)
\(2A=3-P\le3-1=2\)
\(2A\le2\Leftrightarrow A\le1\)
\("="\Leftrightarrow a=b=c\)
10 tháng 10 2021
\(\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{a^2+b^2+c^2}{ab+bc+ca}\\ =2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\dfrac{a^2+b^2+c^2}{ab+bc+ca}\)
ta có: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)(BĐT Nesbit)
\(\Rightarrow2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\ge\dfrac{2.3}{2}=3\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta cm \(\dfrac{a^2+b^2+c^2}{ab+bc+ca}\ge1\)
Thật vậy: \(\dfrac{a^2+b^2+c^2}{ab+bc+ca}\ge1\\ \Rightarrow a^2+b^2+c^2\ge ab+bc+ca\\ \Rightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\\ \Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\\ \Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Dấu "=" xảy ra khi \(a=b=c\)
\(\Rightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{a^2+b^2+c^2}{ab+bc+ca}\ge3+1=4\)
Dấu "=" xảy ra khi \(a=b=c\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 8 2021
\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)
\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự và cộng lại:
\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1
Bunhiacopxki:
\(\left(b+a+a\right)\left(b+c+\dfrac{c^2}{a}\right)\ge\left(b+\sqrt{ca}+c\right)^2\)
\(\Rightarrow\dfrac{2a^2+ab}{\left(b+\sqrt{ca}+c\right)^2}\ge\dfrac{2a^2+ab}{\left(2a+b\right)\left(b+c+\dfrac{c^2}{a}\right)}=\dfrac{a^2}{c^2+ab+bc}\)
Tương tự:
\(\dfrac{2b^2+bc}{\left(c+\sqrt{ca}+a\right)^2}\ge\dfrac{b^2}{a^2+ab+bc}\)
\(\dfrac{2c^2+ca}{\left(a+\sqrt{bc}+b\right)^2}\ge\dfrac{c^2}{b^2+ac+bc}\)
\(\Rightarrow P\ge\dfrac{a^2}{c^2+ab+ac}+\dfrac{b^2}{a^2+ab+bc}+\dfrac{c^2}{b^2+ac+bc}\)
\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)
Dấu "=" xảy ra khi \(a=b=c\)
ĐKXĐ: \(ab+bc+ca\ne0\)
- Nếu 1 biến bằng 0 thì BĐT hiển nhiên đúng
- Nếu cả 3 biến đều khác 0:
\(\Leftrightarrow\dfrac{2a^2}{2a^2+bc}+\dfrac{2b^2}{2b^2+ca}+\dfrac{2c^2}{2c^2+ab}\le2\)
\(\Leftrightarrow\dfrac{bc}{2a^2+bc}+\dfrac{ca}{2b^2+ca}+\dfrac{ab}{2c^2+ab}\ge1\)
Ta có:
\(VT=\dfrac{\left(bc\right)^2}{2a^2bc+\left(bc\right)^2}+\dfrac{\left(ca\right)^2}{2ab^2c+\left(ca\right)^2}+\dfrac{\left(ab\right)^2}{2abc^2+\left(ab\right)^2}\)
\(VT\ge\dfrac{\left(ab+bc+ca\right)^2}{\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)}=\dfrac{\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2}=1\) (đpcm)
Dấu "=" xảy ra khi 3 biến bằng nhau hoặc 1 biến bằng 0, 2 biến bằng nhau