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\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=1\Rightarrow1-\dfrac{1}{1+a}=\dfrac{1}{1+b}+\dfrac{1}{1+c}\)
\(\Rightarrow\dfrac{a}{1+a}\ge\dfrac{1}{1+b}+\dfrac{1}{1+c}\ge2\sqrt{\dfrac{1}{\left(1+b\right)\left(1+c\right)}}\) (1)
Tương tự ta có:
\(\dfrac{b}{1+b}\ge2\sqrt{\dfrac{1}{\left(1+a\right)\left(1+c\right)}}\) (2)
\(\dfrac{c}{1+c}\ge2\sqrt{\dfrac{1}{\left(1+a\right)\left(1+b\right)}}\) (3)
Nhân vế (1);(2);(3):
\(\Rightarrow\dfrac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
\(\Rightarrow abc\ge8\)
Dấu "=" xảy ra khi \(a=b=c=2\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{\left(1+1+1\right)^2}{3+a+b+c+}=\frac{9}{6}=\frac{3}{2}\)
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)
\(=1+\dfrac{a}{b}+\dfrac{a}{c}+1+\dfrac{b}{a}+\dfrac{b}{c}+1+\dfrac{c}{a}+\dfrac{c}{b}\)
\(=3+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}\)\(\ge3+2+2+2=9\)
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 <=> \(a=b=c=\frac{1}{3}\)
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...