Chứng minh: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\) (với a, b > 1)
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TB
1
26 tháng 9 2018
Bạn cần biết \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) (nếu bạn chưa biết thì xét hiệu)
Ta có: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\)
\(\ge\frac{4}{1+a^2+1+b^2}\)
\(=\frac{4}{a^2+b^2+2}\)
\(\ge\frac{4}{2ab+2}=\frac{2}{ab+1}\)
Dấu "=" xảy ra khi \(a=b\)
AS
4
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2019
Với \(a;b;c\ne0\) ta luôn có:
\(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2\ge0\)
\(\Leftrightarrow\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}\ge0\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
Dấu "=" xảy ra khi \(a=b=c\ne0\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2019
Do \(x;y\in N\) *\(\Rightarrow x+y\ge2\)
\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)
\(\Rightarrow\frac{x+y}{x^2+y^2}\le\frac{2\left(x+y\right)}{\left(x+y\right)^2}=\frac{2}{x+y}\)
\(\Rightarrow\frac{2}{x+y}\ge\frac{3}{5}\Rightarrow x+y\le\frac{10}{3}\)
\(\Rightarrow x+y=\left\{2;3\right\}\)
TH1: \(x=y=1\Rightarrow\frac{x+y}{x^2+y^2}=1\left(ktm\right)\)
TH2: \(\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)
\(\Rightarrow\frac{x+y}{x^2+y^2}=\frac{3}{5}\left(tm\right)\)
TN
3
19 tháng 11 2017
Giả sử\(\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge\frac{2}{ab+1}\)
\(\Leftrightarrow\left(\frac{1}{x^2+1}-\frac{1}{xy+1}\right)+\left(\frac{1}{1+y^2}-\frac{1}{xy+1}\right)\ge0\)
\(\Leftrightarrow\frac{1+xy-1-x^2}{\left(1+x^2\right)\left(xy+1\right)}+\frac{1+xy-1-y^2}{\left(1+y^2\right)\left(xy+1\right)}\ge0\)
\(\Leftrightarrow\frac{x\left(y-x\right)}{\left(1+x^2\right)\left(xy+1\right)}+\frac{y\left(x-y\right)}{\left(1+y^2\right)\left(xy+1\right)}\ge0\)
\(\Leftrightarrow\frac{x\left(y-x\right)\left(1+y^2\right)+y\left(x-y\right)\left(1+x^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(xy+1\right)}\ge0\)
\(\Leftrightarrow\left(x-y\right)\left[-x\left(1+y^2\right)+y\left(1+x^2\right)\right]\ge0\) Do x;y>1
\(\Leftrightarrow\left(x-y\right)^2\left(xy-1\right)\ge0\) (BĐT đúng do x;y>1)
Vậy............
19 tháng 11 2017
chuyển vế qua biến đổi tương đương tách 2/1+ab ra là 1/1+ab +1/1+ab
4 tháng 3 2020
BĐT tương đương :
\(\frac{1}{1+x^2}-\frac{1}{1+xy}+\frac{1}{1+y^2}-\frac{1}{1+xy}\ge0\)
\(\Leftrightarrow\frac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\) ( đúng )
Dấu "=" xảy ra \(\Leftrightarrow a=b=1\)
NT
2
KN
13 tháng 4 2020
\(\frac{1}{1+a}+\frac{1}{1+b}\ge\frac{2}{1+\sqrt{ab}}\Leftrightarrow\frac{1}{1+a}+\frac{1}{1+b}-\frac{2}{1+\sqrt{ab}}\ge0\)
\(\Leftrightarrow\left(\frac{1}{a+1}-\frac{1}{1+\sqrt{ab}}\right)+\left(\frac{1}{b+1}-\frac{1}{1+\sqrt{ab}}\right)\ge0\)
\(\Leftrightarrow\frac{\sqrt{ab}-a}{\left(a+1\right)\left(1+\sqrt{ab}\right)}+\frac{\sqrt{ab}-b}{\left(b+1\right)\left(1+\sqrt{ab}\right)}\ge0\)
\(\Leftrightarrow\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\left(a+1\right)\left(1+\sqrt{ab}\right)}+\frac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(b+1\right)\left(1+\sqrt{ab}\right)}\ge0\)
\(\Leftrightarrow\frac{-\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)\left(b+1\right)+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\left(a+1\right)}{\left(a+1\right)\left(b+1\right)\left(1+\sqrt{ab}\right)}\ge0\)
\(\Leftrightarrow\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a\sqrt{b}+\sqrt{b}-b\sqrt{a}-\sqrt{a}\right)}{\left(a+1\right)\left(b+1\right)\left(1+\sqrt{ab}\right)}\ge0\)
\(\Leftrightarrow\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{ab}-1\right)}{\left(a+1\right)\left(b+1\right)\left(1+\sqrt{ab}\right)}\ge0\)(đúng với \(ab\ge1\))
Vậy \(\frac{1}{1+a}+\frac{1}{1+b}\ge\frac{2}{1+\sqrt{ab}}\)
Đẳng thức xảy ra khi a = b
LQ
0
ML
1
9 tháng 7 2019
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{ab+1}\)
\(\Leftrightarrow\frac{1}{a^2+1}-\frac{1}{ab+1}+\frac{1}{b^2+1}-\frac{1}{ab+1}\ge0\)
\(\Leftrightarrow\frac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\frac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\frac{\left(ab-a^2\right)\left(b^2+1\right)+\left(ab-b^2\right)\left(a^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\frac{-a\left(b^2+1\right)\left(a-b\right)+b\left(a-b\right)\left(a^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)\left(-ab^2-a+a^2b+b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)\left[ab\left(a-b\right)-\left(a-b\right)\right]}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)( luôn đúng )
Dấu "=" xảy ra \(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\ab-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\\ab=1\end{matrix}\right.\)
22 tháng 6 2021
vì \(a+b+c=1\)
\(< =>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)
\(=3+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}+\frac{a}{c}\)
\(=3+\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\)
ta có pt:
\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(3+\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\right)\)
\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{3}{4}+\frac{a^2+b^2}{4ab}+\frac{b^2+c^2}{4bc}+\frac{c^2+a^2}{4ca}\)
áp dụng bđt cô- si( cauchy) gọi pt là P
\(P\ge2\sqrt{\frac{ab}{a^2+b^2}\frac{a^2+b^2}{4ab}}+2\sqrt{\frac{bc}{b^2+c^2}\frac{b^2+c^2}{4bc}}+2\sqrt{\frac{ca}{c^2+a^2}\frac{c^2+a^2}{4ca}}+\frac{3}{4}\)
\(P\ge2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+\frac{3}{4}\)
\(P\ge2.\frac{1}{2}+2.\frac{1}{2}+2.\frac{1}{2}+\frac{3}{4}\)
\(P\ge1+1+1+\frac{3}{4}=\frac{15}{4}\)
dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)
<=>ĐPCM
(1 + a^2 ) + 1/(1+b^2) \(\ge\) 2/( 1+ab)
<=> (1+ b^2)(1+ab) + (1+a^2)(1 +ab) \(\ge\) 2(1+a^2)(1+ b^2)
<=>1 + b^2 +ab + ab^3 + 1 +a^2 +ab + a^3b - 2(1 +a^2 +b^2 +a^2b^2) \(\ge\) 0
<=> ab(a^2 - 2ab +b^2) - (a^2 +2ab +b^2) \(\ge\) 0
<=> (ab -1)(a-b)^2 \(\ge\) 0
Điều này hiển nhiên đúng do ab \(\ge\) 1; (a-b)^2 \(\ge\) 0
Dấu "=" khi và chỉ khi a =b =1