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Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{\left(1+1+1\right)^2}{3+a+b+c+}=\frac{9}{6}=\frac{3}{2}\)
c)\(\frac{a^2}{b^2}+\frac{b^2}{a^2}+4\ge3\cdot\left(\frac{a}{b}+\frac{b}{a}\right)\)
Thế : \(\frac{\left(a-b\right)^2\left(a^2-ab+b^2\right)}{a^2b^2}\ge0\)
\(\Leftrightarrow\frac{\left(b-a\right)^2\left(a^2-ab+b^2\right)}{a^2b^2}\ge0\)
\(\Leftrightarrow\frac{a^4+4a^2b^2+b^4}{a^2b^2}\ge\frac{3\left(a^2+b^2\right)}{ab}\)
\(\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{a^2}+4\ge\frac{3a}{b}+\frac{3b}{a}\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
\(\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{a^2}+4>=3\cdot\left(\frac{a}{b}+\frac{b}{a}\right)\)
(a+b+c)3= (a+b)3+3(a+b)2c+3(a+b)c2+c2
=a3+3a2b+3ab2+b2+3(a+b)c(a+b+c)+c2
=a3+b3+c3+3ab(a+b)+3(a+b)c(a+b+c)
=a3+b3+c3+3(a+b)[ab+c(a+b+c)]
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
=a3+b3+c3+3(a+b)[(ab+ac)+(bc+c2)]
=a3+b3+c3+3(a+b)[a(b+c)+c(b+c)]
=a3+b3+c3+3(a+b)(b+c)(c+a)
Vậy (a+b+c)3 = a3 + b3 + c3 + 3(a+b)(b+c)(c+a)
Áp dụng bất đẳng thức Cauchy - Schwarz dưới dạng Engel ta có :
\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{1+1+1}=\frac{1}{3}\) (đpcm)
Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{3}\)
1, a=ƯCLN(128;48;192)
2, b= ƯCLN(300;276;252)
3, Gọi n.k+11=311 => n.k = 300
n.x + 13 = 289 => n.x = 276
=> \(n\inƯC\left(300;276\right)\)
4, G/s (2n+1;6n+5) = d (d tự nhiên)
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\6n+5⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\6n+5⋮d\end{cases}}}\) \(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+5⋮d\end{cases}\Rightarrow6n+5-\left(6n+3\right)⋮d}\)
\(\Rightarrow2⋮d\Rightarrow d\in\left\{1;2\right\}\)
Vì 2n+1 lẻ => 2n+1 không chia hết cho 2
=> d khác 2 => d=1 => đpcm
Với a,b,c > 0 áp dụng BĐT Cauchy, ta có
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
Cmtt: \(\dfrac{c}{a}+\dfrac{a}{c}\ge2\) và \(\dfrac{b}{c}+\dfrac{c}{b}\ge2\)
Theo đề bài, ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)(do a + b + c = 1)
\(=1+\dfrac{a}{b}+\dfrac{a}{c}+1+\dfrac{b}{a}+\dfrac{b}{c}+1+\dfrac{c}{a}+\dfrac{c}{b}\)
\(=3+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}\)\(\ge3+2+2+2=9\)
a) Ta có:
\(5^2=25\equiv-1\left(mod13\right)\)
\(\Rightarrow\left\{{}\begin{matrix}5^{2004}=\left(5^2\right)^{1002}\equiv\left(-1\right)^{1002}\left(mod13\right)\equiv1\left(mod13\right)\\5^{2002}=\left(5^2\right)^{1001}\equiv\left(-1\right)^{1001}\left(mod13\right)\equiv-1\left(mod13\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5^{2005}=5^{2004}.5\equiv1.5\left(mod13\right)\equiv5\left(mod13\right)\\5^{2003}=5^{2002}.5\equiv\left(-1\right).5\left(mod13\right)\equiv-5\left(mod13\right)\end{matrix}\right.\)
\(\Rightarrow5^{2005}+5^{2003}\equiv5+\left(-5\right)\left(mod13\right)\equiv0\left(mod13\right)\)
Vậy...
mod?