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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 <=> \(a=b=c=\sqrt{3}\)
N
0
TN
19 tháng 5 2017
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(3P=\frac{3\sqrt{ab}}{c+3\sqrt{bc}}+\frac{3\sqrt{bc}}{a+3\sqrt{bc}}+\frac{3\sqrt{ca}}{b+3\sqrt{ca}}\)
\(=3-\left(\frac{a}{a+3\sqrt{bc}}+\frac{b}{b+3\sqrt{ca}}+\frac{c}{c+3\sqrt{ab}}\right)\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right]\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{\left(a^2+b^2+c^2\right)+3\left(ab+bc+ca\right)}\right]\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{\left(a^2+b^2+c^2\right)+\frac{\left(a+b+c\right)^2}{3}}\right]=3-\frac{9}{4}=\frac{3}{4}\)
Đẳng thức xảy ra khi \(a=b=c\)
20 tháng 5 2017
Bạn Thắng Nguyễn chạy mà bị kiệt sức ở giai đoạn cuối để mình chạy tiếp sức phần còn lại nhé.
Từ \(3-\left[\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right]\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{\left(a+b+c\right)^2}{3}}\right]=3-\frac{3}{4}=\frac{9}{4}\)
\(\Rightarrow P\le\frac{\frac{9}{4}}{3}=\frac{3}{4}\)
T
18 tháng 2 2020
Có: \(9=\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow3\ge ab+bc+ca\)
Từ đây: \(D=\Sigma_{cyc}\frac{ab}{\sqrt{c^2+3}}\le\Sigma_{cyc}\frac{ab}{\sqrt{c^2+ab+bc+ca}}\)
\(=\Sigma_{cyc}\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=\Sigma_{cyc}\sqrt{\frac{ab}{a+c}}.\sqrt{\frac{ab}{b+c}}\le\Sigma_{cyc}\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
\(=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
18 tháng 11 2017
a) Gõ link này nha: http://olm.vn/hoi-dap/question/1078496.html
NY
3 tháng 2 2020
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
PG
3 tháng 2 2020
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
\(P=\frac{\sqrt{ab}}{c+3\sqrt{ab}}+\frac{\sqrt{bc}}{a+3\sqrt{bc}}+\frac{\sqrt{ca}}{b+3\sqrt{ca}}\)
\(=1-\frac{1}{3}\left(\frac{c^2}{c^2+3c\sqrt{ab}}+\frac{a^2}{a^2+3a\sqrt{bc}}+\frac{b^2}{b^2+3b\sqrt{ca}}\right)\)
\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right)\)
\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{\left(a+b+c\right)^2}{3}}\right)\)
\(\le1-\frac{1}{3}\cdot\frac{\left(a+b+c\right)^2}{\frac{4\left(a+b+c\right)^2}{3}}=1-\frac{1}{4}=\frac{3}{4}\)
\(P=\frac{\sqrt{ab}}{c+3\sqrt{ab}}+\frac{\sqrt{bc}}{a+3\sqrt{bc}}+\frac{\sqrt{ca}}{b+3\sqrt{ca}}\)
\(=1-\frac{1}{3}\left(\frac{c^2}{c^2+3c\sqrt{ab}}+\frac{a^2}{a^2+3a\sqrt{bc}}+\frac{b^2}{b^2+3b\sqrt{ca}}\right)\)
\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right)\)
\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)
\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{\left(a+b+c\right)^2}{3}}\right)\)
\(\le1-\frac{1}{3}\cdot\frac{\left(a+b+c\right)^2}{\frac{4\left(a+b+c\right)^2}{3}}\)
\(=1-\frac{1}{4}\)
\(=\frac{3}{4}\)