cho a,b,c> 0 thỏa mãn ab+bc+ca =3. Cmr:
\(\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}\ge\frac{3}{4}\)
K
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N
4 tháng 7 2020
\(\frac{a^3}{b^2+3}=\frac{a^3}{b^2+ab+bc+ca}=\frac{a^3}{\left(a+b\right)\left(b+c\right)}\)
Tương tự
\(\Rightarrow\Sigma_{cyc}\frac{a^3}{b^2+3}=\Sigma_{cyc}\frac{a^3}{\left(a+b\right)\left(b+c\right)}\)
Theo Cô-si:\(\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{a+b}{8}+\frac{b+c}{8}\ge\frac{3}{4}a\)
\(\Rightarrow\Sigma_{cyc}\frac{a^3}{\left(a+b\right)\left(b+c\right)}\ge\frac{1}{4}\left(a+b+c\right)\ge\frac{1}{4}\sqrt{3\left(ab+bc+ca\right)}=\frac{3}{4}\)
HM
6 tháng 4 2018
Cho mk k nhé!
4/1x3x5 = 1/1x3 - 1/3x5
4/3x5x7 = 1/3x5 - 1/5x7
.............
A = 1/1x3 - 1/11x13
1/1x3x5 = 1/4 x (1/1x3 - 1/3x5)
1/3x5x7 = 1/4 x (1/3x5 - 1/5x7)
..........
B = 1/4 x (1/1x3 - 1/11x13)
AH
Akai Haruma
Giáo viên
18 tháng 8 2019
Lời giải:
Ta có:
\(\text{VT}=\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}=\frac{a^3}{b^2+ab+bc+ac}+\frac{b^3}{c^2+ab+bc+ac}+\frac{c^3}{a^2+ab+bc+ac}\)
\(=\frac{a^3}{(b+a)(b+c)}+\frac{b^3}{(c+a)(c+b)}+\frac{c^3}{(a+b)(a+c)}\)
Áp dụng BĐT AM-GM:
\(\frac{a^3}{(b+a)(b+c)}+\frac{b+a}{8}+\frac{b+c}{8}\geq 3\sqrt[3]{\frac{a^3}{8.8}}=\frac{3a}{4}\)
\(\frac{b^3}{(c+a)(c+b)}+\frac{c+a}{8}+\frac{c+b}{8}\geq \frac{3b}{4}\)
\(\frac{c^3}{(a+b)(a+c)}+\frac{a+b}{8}+\frac{a+c}{8}\geq \frac{3c}{4}\)
Cộng theo vế và rút gọn thu được:
\(\text{VT}\geq \frac{a+b+c}{4}\)
Tiếp tục áp dụng BĐT AM-GM: \((a+b+c)^2\geq 3(ab+bc+ac)=9\Rightarrow a+b+c\geq 3\)
Do đó: \(\text{VT}\geq \frac{3}{4}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$
8 tháng 1 2020
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
28 tháng 8 2019
\(sigma\frac{a^2+b^2}{ab\left(a+b\right)^3}\ge sigma\frac{\frac{\left(a+b\right)^2}{2}}{\left(a+b\right)^2\left(a^3+b^3\right)}=sigma\frac{1}{2\left(a^3+b^3\right)}\ge\frac{9}{4\left(a^3+b^3+c^3\right)}=\frac{9}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 9 2019
\(3=ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\Rightarrow a+b+c\ge3\)
Ta có: \(\frac{a^3}{b^2+3}=\frac{a^3}{b^2+ab+bc+ca}=\frac{a^3}{\left(a+b\right)\left(b+c\right)}\)
Mặt khác \(\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{a+b}{8}+\frac{b+c}{8}\ge\frac{3a}{4}\)
Tương tự: \(\frac{b^3}{c^3+3}+\frac{a+c}{8}+\frac{b+c}{8}\ge\frac{3b}{4}\) ; \(\frac{c^3}{a^2+8}+\frac{a+b}{8}+\frac{a+c}{8}\ge\frac{3c}{4}\)
Cộng vế với vế:
\(P+\frac{1}{2}\left(a+b+c\right)\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow P\ge\frac{1}{4}\left(a+b+c\right)\ge\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
9 tháng 8 2020
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
7 tháng 7 2018
Từ giả thiết:\(ab+bc+ca=3\Rightarrow\left(ab+bc+ca\right)^2=9\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=9\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=9-2abc\left(a+b+c\right)\)
Ta có:\(\frac{a}{2a^2+bc}+\frac{b}{2b^2+ca}+\frac{c}{2c^2+ab}\)\(=\frac{1}{\frac{2a^2+bc}{a}}+\frac{1}{\frac{2b^2+ca}{b}}+\frac{1}{\frac{2c^2+ab}{c}}\)
\(\ge\frac{\left(1+1+1\right)^2}{2a+\frac{bc}{a}+2b+\frac{ca}{b}+2c+\frac{ab}{c}}=\frac{9}{2a+2b+2c+\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}}\)
\(=\frac{9}{2a+2b+2c+\frac{b^2c^2+c^2a^2+a^2b^2}{abc}}=\frac{9}{2a+2b+2c+\frac{9-2abc\left(a+b+c\right)}{abc}}\)
\(=\frac{9}{2a+2b+2c+\frac{9}{abc}-2\left(a+b+c\right)}=\frac{9}{\frac{9}{abc}}=abc\)
Dấu "=" xảy ra khi
\(\frac{2a^2+bc}{a}=\frac{2b^2+ca}{b}=\frac{2c^2+ab}{c}=\frac{2a^2+bc-2b^2-ca}{a-b}=\frac{2\left(a-b\right)\left(a+b\right)-c\left(a-b\right)}{a-b}\)
\(=2\left(a+b\right)-c\).Tương tự ta có:\(2\left(a+b\right)-c=2\left(b+c\right)-a=2\left(c+a\right)-b\)
\(\Leftrightarrow a+b=b+c=c+a\)
\(\Leftrightarrow a=b=c\)
\(VT=\frac{a^3}{b^2+8}+\frac{b^3}{c^2+8}+\frac{c^3}{a^2+8}\)
\(=\frac{a^3}{b^2+ab+bc+ca}+\frac{b^3}{c^2+ab+bc+ca}+\frac{c^3}{a^2+ab+bc+ca}\)
\(=\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{b^3}{\left(b+c\right)\left(c+a\right)}+\frac{c^3}{\left(c+a\right)\left(a+b\right)}\)
Áp dụng BĐT Cô si ta có :
\(\left\{{}\begin{matrix}\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{a+b}{8}+\frac{b+c}{8}\ge\frac{3a}{4}\\\frac{b^3}{\left(b+c\right)\left(c+a\right)}+\frac{b+c}{8}+\frac{c+a}{8}\ge\frac{3b}{4}\\\frac{c^3}{\left(c+a\right)\left(a+b\right)}+\frac{c+a}{8}+\frac{a+b}{8}\ge\frac{3c}{4}\end{matrix}\right.\)
\(\Rightarrow\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{b^3}{\left(b+c\right)\left(c+a\right)}+\frac{c^3}{\left(c+a\right)\left(a+b\right)}\ge\frac{a+b+c}{4}\ge\frac{\sqrt{3\left(ab+bc+ca\right)}}{4}=\frac{3}{4}\)
Vậy BĐT được chứng minh . Dấu = xảy ra khi \(a=b=c=1\)