\(Cho a,b,c>0. Cmr: \dfrac{a^3b}{1+ab^2}+\dfrac{b^3c}{1+bc^2}+\dfrac{c^3a}{1+ca^2}>\dfrac{abc(a+b+c)}{1+abc}\)
K
Khách
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13 tháng 6 2021
Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)
CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)
\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)
Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu = xảy ra khi a=b=c=3
13 tháng 6 2021
Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)
\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)
Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)
\(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)
\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)
Vậy...
21 tháng 5 2022
Ta có BĐT: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=3.3=9\)
\(\Rightarrow a+b+c\ge3\)
Phân tích và áp dụng BĐT AM-GM:
\(\dfrac{1+3a}{1+b^2}=\dfrac{1}{1+b^2}+\dfrac{3a}{1+b^2}=\left(1-\dfrac{b^2}{1+b^2}\right)+\left(3a-\dfrac{3ab^2}{1+b^2}\right)\ge\left(1-\dfrac{b^2}{2b}\right)+\left(3a-\dfrac{3ab^2}{2b}\right)=\left(1-\dfrac{b}{2}\right)+\left(3a-\dfrac{3}{2}ab\right)\)
Tương tự:
\(\dfrac{1+3b}{1+c^2}\ge\left(1-\dfrac{c}{2}\right)+\left(3b-\dfrac{3}{2}bc\right)\)
\(\dfrac{1+3c}{1+a^2}\ge\left(1-\dfrac{a}{2}\right)+\left(3c-\dfrac{3}{2}ca\right)\)
Cộng các vế của các BĐT ta được:
\(P\ge3-\dfrac{1}{2}\left(a+b+c\right)+3\left(a+b+c\right)-\dfrac{3}{2}\left(ab+bc+ca\right)=3+\dfrac{5}{2}\left(a+b+c\right)-\dfrac{3}{2}.3\ge3+\dfrac{5}{2}.3-\dfrac{9}{2}=6\)
\(P=6\Leftrightarrow a=b=c=1\)
Vậy \(P_{min}=6\)
KK
12 tháng 5 2017
a) Áp dụng bất đẳng thức Schur với \(r=1\)
\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)
\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)
\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
b) Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
c) Ta có \(abc=ab+bc+ca\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)
\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\dfrac{3}{16}\)
\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )
23 tháng 3 2018
Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
24 tháng 3 2018
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 1
\(ab+bc+ca=3abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x;y;z>0\\x+y+z=3\end{matrix}\right.\)
\(P=\dfrac{x}{\left(3-x\right)^2}+\dfrac{y}{\left(3-y\right)^2}+\dfrac{z}{\left(3-z\right)^2}\)
Ta có đánh giá sau: \(\dfrac{t}{\left(3-t\right)^2}\ge\dfrac{2t-1}{4};\forall t\in\left(0;3\right)\)
Thực vậy, BĐT đã cho tương đương:
\(4t\ge\left(2t-1\right)\left(3-t\right)^2\)
\(\Leftrightarrow-2t^3+13t^2-20t+9\ge0\)
\(\Leftrightarrow\left(9-2t\right)\left(t-1\right)^2\ge0\) (luôn đúng với \(t< 3\))
Áp dụng ta được:
\(P\ge\dfrac{2x-1}{4}+\dfrac{2y-1}{4}+\dfrac{2z-1}{4}=\dfrac{2\left(x+y+z\right)-3}{4}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 1
Cách khác:
Sau khi đặt ẩn phụ, ta có:
\(P=\dfrac{x}{\left(3-x\right)^2}+\dfrac{y}{\left(3-y\right)^2}+\dfrac{z}{\left(3-z\right)^2}=\dfrac{x}{\left(y+z\right)^2}+\dfrac{y}{\left(z+x\right)^2}+\dfrac{z}{\left(x+y\right)^2}\)
\(\Rightarrow3P=\left(x+y+z\right)\left(\dfrac{x}{\left(y+z\right)^2}+\dfrac{y}{\left(z+x\right)^2}+\dfrac{z}{\left(x+y\right)^2}\right)\ge\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)^2\ge\dfrac{9}{4}\)
(BĐT Netsbitt)
\(\Rightarrow P\ge\dfrac{3}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 4 2023
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Tương tự:
\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{c+a}+\dfrac{b}{2}\right)\)
\(\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)
Cộng vế:
\(VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ca+ab}{b+c}+\dfrac{bc+ab}{c+a}+\dfrac{a+b+c}{2}\right)=\dfrac{a+b+c}{6}\)
Dấu "=" xảy ra khi \(a=b=c\)
NV
1 tháng 4 2023
\(ab+bc+ca=3abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\) (do a,b,c là các số dương)
Áp dụng BĐT Bunhiacopxki dạng phân thức:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{6^2}{a+2b+3c}\)
\(\Rightarrow\dfrac{36}{a+2b+3c}\le\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}\left(1\right)\)
Tương tự: \(\left\{{}\begin{matrix}\dfrac{36}{b+2c+3a}\le\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{3}{a}\left(2\right)\\\dfrac{36}{c+2a+3b}\le\dfrac{1}{c}+\dfrac{2}{a}+\dfrac{3}{b}\left(3\right)\end{matrix}\right.\)
Lấy (1) + (2) + (3) ta được:
\(36F\le6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=6.3=18\)
\(\Rightarrow F\le\dfrac{1}{2}\)
MaxF=1/2 khi \(a=b=c=1\)
\(VT=\dfrac{a^3bc}{c+ab^2c}+\dfrac{ab^3c}{a+abc^2}+\dfrac{abc^3}{b+a^2bc}\)
\(=abc\left(\dfrac{a^2}{c+ab^2c}+\dfrac{b^2}{a+abc^2}+\dfrac{c^2}{b+a^2bc}\right)\)
Áp dụng bđt Cauchy-Schwarz dạng engel có:
\(VT\ge\dfrac{abc\left(a+b+c\right)^2}{a+b+c+abc\left(a+b+c\right)}\)\(=\dfrac{abc\left(a+b+c\right)}{1+abc}\)
Dấu "=" xảy ra khi \(a=b=c\)
Vậy...
Sai đề không bạn,tại a=b=c=2 thay vào không thỏa mãn nha