Cho a>0, b>0,m>0,n>0 (m>n)
c/m\(\dfrac{a}{na+mb}+\dfrac{b}{nb+ma}\ge\dfrac{2}{m+n}\)
K
Khách
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ND
29 tháng 4 2018
Bài 1:
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\) với a,b,c > 0
Áp dụng BĐT Chauchy cho 2 số không âm, ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge c\sqrt{\dfrac{b}{a}.\dfrac{a}{b}}=2c\)
\(\dfrac{ac}{b}+\dfrac{ab}{c}=a\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\ge a\sqrt{\dfrac{c}{b}.\dfrac{b}{c}}=2a\)
\(\dfrac{ab}{c}+\dfrac{bc}{a}=b\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge b\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2b\)
Cộng vế theo vế ta được:
\(2\left(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
T
31 tháng 7 2017
1. Vì x, y, z > 0
\(xy+yz+zx\ge2xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge2\)
Suy ra:
\(\dfrac{1}{x}\ge1-\dfrac{1}{y}+1-\dfrac{1}{z}=\dfrac{y-1}{y}+\dfrac{z-1}{z}\ge2\sqrt{\dfrac{\left(y-1\right)\left(z-1\right)}{yz}}\). (1)
Tương tự \(\dfrac{1}{y}\ge2\sqrt{\dfrac{\left(z-1\right)\left(x-1\right)}{zx}}\) (2)
và \(\dfrac{1}{z}\ge2\sqrt{\dfrac{\left(x-1\right)\left(y-1\right)}{xy}}\) (3)
Nhân (1), (2), (3) với nhau theo vế ta được
\(\dfrac{1}{xyz}\ge\dfrac{8\left(x-1\right)\left(y-1\right)\left(z-1\right)}{xyz}\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)\le\dfrac{1}{8}\)
Đẳng thức xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{2}\)
N
7 tháng 1 2018
\(VT\ge a+b+c+\dfrac{9}{2\left(ab+bc+ca\right)}\ge\sqrt{3\left(ab+bc+ca\right)}+\dfrac{9}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{9}{2\left(ab+bc+ca\right)}\ge3\sqrt[3]{\dfrac{27}{8}}=\dfrac{9}{2}\)
7 tháng 1 2018
Áp dụng BĐT Cauchy ta có
\(\dfrac{b^2}{a}+a\ge2b;\) \(\dfrac{c^2}{b}+b\ge2c\); \(\dfrac{a^2}{c}+c\ge2a\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}\ge a+b+c\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{9}{2\left(ab+bc+ac\right)}\ge a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\)Ta phải chứng minh
\(a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow4\left(a+b+c\right)\left(ab+bc+ac\right)+18\ge18\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge0\)
Áp dụng BĐT Cauchy:
\(ab+bc+ac\ge3\sqrt[3]{a^2b^2c^2}=3\)
\(a+b+c\ge3\sqrt[3]{abc}=3\)
\(\Rightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge3\left(4.3-18\right)+18=0\)=> đpcm
KK
2 tháng 4 2017
\(VT=a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{ca^2}{a^2+1}\)
\(VT=3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}b^2+1\ge2\sqrt{b^2}=2b\\c^2+1\ge2\sqrt{c^2}=2c\\a^2+1\ge2\sqrt{a^2}=2a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\\\dfrac{bc^2}{c^2+1}\le\dfrac{bc^2}{2c}=\dfrac{bc}{2}\\\dfrac{ca^2}{a^2+1}\le\dfrac{ca^2}{2a}=\dfrac{ca}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge3-\dfrac{ab+bc+ca}{2}\) ( 1 )
Theo hệ quả của bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow3\ge ab+bc+ca\)
\(\Rightarrow\dfrac{3}{2}\ge\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\) ( 2 )
Từ (1) và (2)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
17 tháng 9 2017
Áp dụng BĐT Cô si Ta có : \(\dfrac{a}{b^2+1}=a-\dfrac{ab^2}{b^2+1}\ge a-\dfrac{ab^2}{2b}=a-\dfrac{ab}{2}\)
\(\dfrac{b}{c^2+1}=b-\dfrac{c^2b}{c^2+1}\ge b-\dfrac{c^2b}{2c}=b-\dfrac{cb}{2}\)
\(\dfrac{c}{a^2+1}=c-\dfrac{a^2c}{a^2+1}\ge c-\dfrac{a^2c}{2a}=c-\dfrac{ac}{2}\)
Cộng ba vế BĐT lại ta được:
\(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge a+b+c-\left(\dfrac{ab+bc+ac}{2}\right)\)
Ta có đánh giá quen thuộc \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{9}{3}=3\)
\(\Rightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)(ĐPCM)
21 tháng 6 2017
làm rõ \(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)
\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)
\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\) (đúng)
21 tháng 6 2017
ok thỏa thuận rồi tui làm nửa sau thui nhé :D
Đặt \(a^2=x;b^2=y;c^2=z\) thì ta có:
\(VT=\sqrt{\dfrac{x}{x+y}}+\sqrt{\dfrac{y}{y+z}}+\sqrt{\dfrac{z}{x+z}}\)
Lại có: \(\sqrt{\dfrac{x}{x+y}}=\sqrt{\dfrac{x}{\left(x+y\right)\left(x+z\right)}\cdot\sqrt{x+z}}\)
Tương tự cộng theo vế rồi áp dụng BĐT C-S ta có:
\(VT^2\le2\left(x+y+z\right)\left[\dfrac{x}{\left(x+y\right)\left(x+z\right)}+\dfrac{y}{\left(y+z\right)\left(y+x\right)}+\dfrac{z}{\left(z+x\right)\left(z+y\right)}\right]\)
\(\Leftrightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
Vì \(VP^2=\dfrac{9}{2}\) nên cần cm \(VT\le \frac{9}{2}\)
\(\Leftrightarrow9\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left(x+y+z\right)\left(xy+yz+xz\right)\)
Can you continue
28 tháng 1 2018
Đặt \(\left\{{}\begin{matrix}\sqrt{a^2+b^2}=x\\\sqrt{b^2+c^2}=y\\\sqrt{c^2+a^2}=z\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{x^2+z^2-y^2}{2}\\b^2=\dfrac{x^2+y^2-z^2}{2}\\c^2=\dfrac{y^2+z^2-x^2}{2}\\x+y+z=\sqrt{2011}\end{matrix}\right.\)
Và \(\left\{{}\begin{matrix}b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}y\\a+b\le\sqrt{2}x\\c+a\le\sqrt{2}z\end{matrix}\right.\)
\(VT=\dfrac{1}{2\sqrt{2}}\left(\dfrac{x^2+z^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}+\dfrac{y^2+z^2-x^2}{x}\right)\)
\(\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{2\left(x+y+z\right)^2}{\left(x+y+z\right)}-\left(x+y+z\right)\right)\)
\(=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{\sqrt{2011}}{2\sqrt{2}}=VP\)
21 tháng 2 2018
a) \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Leftrightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\left(đúng\forall a;b\right)\)
Vậy bdt đã được cm
b) \(K=n\left(n+1\right)\left(n+2\right)\left(n+3\right)=\left(n^2+3n\right)\left(n^2+3n+2\right)\)
\(=\left(n^2+3n\right)^2+2\left(n^2+3n\right)\)
Ta có :
\(\left(n^2+3n\right)^2< \left(n^2+3n\right)^2+2\left(n^2+3n\right)< \left(n^2+3n\right)^2+2\left(n^2+3n\right)+1\)
\(\Leftrightarrow\left(n^2+3n\right)^2< \left(n^2+3n\right)^2+2\left(n^2+3n\right)< \left(n^2+3n+1\right)^2\)
Mà \(n^2+3n;n^2+3n+1\) là 2 số tn liên tiếp
\(\Rightarrow K\) không phải số chính phương
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a}{na+mb}+\dfrac{b}{nb+ma}\)
\(=\dfrac{a^2}{na^2+mab}+\dfrac{b^2}{nb^2+mab}\)
\(\ge\dfrac{\left(a+b\right)^2}{na^2+nb^2+2mab}\). Cần chứng minh BĐT
\(\dfrac{\left(a+b\right)^2}{na^2+nb^2+2mab}\ge\dfrac{2}{m+n}\)
Điều này đúng vì tương đương với \(\left(a-b\right)^2\left(m-n\right)\ge0\forall a,b,m,n>0;m>n\)