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\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
=\(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
=\(\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
=\(\dfrac{1}{2}-\dfrac{1}{4n+2}< \dfrac{1}{2}\)
đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
=> 2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+......+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
<=> 2A=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{2n-2}-\dfrac{1}{2n+1}\)
<=>2A=\(1-\dfrac{1}{2n+1}\)
<=> A=\(\left(1-\dfrac{1}{2n+1}\right)\)\(.\dfrac{1}{2}\)
<=> A=\(\dfrac{1}{2}-\dfrac{1}{2\left(2n+1\right)}\)
=>\(A< \dfrac{1}{2}\) (đpcm)
\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\) \(\Rightarrow2S=1-\dfrac{1}{2n+1}\)
\(\Rightarrow S=\dfrac{n}{2n+1}\)
Ta có : \(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
ta được \(\dfrac{1}{1.3}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right);\dfrac{1}{3.5}=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right);\dfrac{1}{5.7}=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\)
\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\) vậy \(S=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{n}{2n+1}\)
Ta có : \(a^2+b^2\ge2ab\Rightarrow a^2+b^2-ab\ge ab\)
\(\Rightarrow\dfrac{1}{a^2-ab+b^2}\le\dfrac{1}{ab}=\dfrac{abc}{ab}=c\) ( do $abc=1$ )
Tương tự ta có :
\(\dfrac{1}{b^2-bc+c^2}\le a\)
\(\dfrac{1}{c^2-ab+a^2}\le b\)
Cộng vế với vế các BĐT trên có :
\(\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{b^2-bc+c^2}+\dfrac{1}{c^2-ac+a^2}\le a+b+c\)
Dấu "=" xảy ra khi $a=b=c$
\(VT=\dfrac{1}{a^2+b^2-ab}+\dfrac{1}{b^2+c^2-bc}+\dfrac{1}{c^2+a^2-ca}\)
\(VT\le\dfrac{1}{2ab-ab}+\dfrac{1}{2bc-bc}+\dfrac{1}{2ca-ca}=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{a+b+c}{abc}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c=1\)
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}\)
\(=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
Do: \(a^2+b^2+c^2=1\text{ nen }a^2\le1,b^2\le1,c^2\le1\)
\(\Rightarrow a\ge-1;b\ge-1;c\ge-1\)
\(\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge0\)
\(\Rightarrow1+a+b+c+ab+bc+ca+abc\ge0\)
Cần C/m:
\(1+a+b+c+ab+bc+ca\ge0\)
Ta có:
\(1+a+b+c+ab+bc+ca\ge0\)
\(\Leftrightarrow a^2+b^2+c^2+ab+bc+ca+a+b+c\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2+2\left(a+b+c\right)+2ab+2bc+2ca+abc\ge0\)
\(\Leftrightarrow\left(a+b+c\right)^2+2\left(a+b+c\right)+1\ge0\)
\(\Leftrightarrow\left(a+b+c+1\right)^2\ge0\left(\text{luon dung}\right)\)
=> ĐPCM
Áp dụng Bđt Cauchy-Schwarz dạng engel ta có:
\(\frac{1}{ac}+\frac{1}{bc}\ge\frac{\left(1+1\right)^2}{ab+bc}=\frac{4}{c\left(a+b\right)}\ge\frac{4}{\frac{\left(a+b+c\right)^2}{4}}=16\)
=>Đpcm
ta có (a-1)2 ≥ 0 ∀a
<=> a2-2a+1 ≥ 0
<=>a2+4a-2a+1 ≥ 4a (cộng cả 2 vế va 4a)
<=> a2+2a+1 ≥ 4a
<=> (a+1)2 ≥ 4a
CM tương tự ta đc
(b+1)2 ≥ 4b
(c+1)2 ≥ 4c
Nhân các vế với nhau ta có
[(a+1)2+(b+1)2 +(c+1)2 ]2 ≥ 4a.4b.4c
<=> [(a+1)2+(b+1)2 +(c+1)2 ]2 ≥64abc
<=> [(a+1)2+(b+1)2 +(c+1)2 ]2 ≥64 (vì abc =1)
<=> (a+1)2+(b+1)2 +(c+1)2 ≥8 (đpcm)