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\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}<\frac{a+c}{a+b+c}+\frac{b+a}{a+b+c}+\frac{c+b}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(1<\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)<2\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\notin N\)
\(\RightarrowĐPCM\)
Sửa đề: chứng minh \(S\ge6\)
Ta có:
\(S=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\left(\frac{a}{b}-2+\frac{b}{a}\right)+\left(\frac{b}{c}-2+\frac{c}{b}\right)+\left(\frac{a}{c}-2+\frac{c}{a}\right)+6\)
\(=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2+\left(\sqrt{\frac{b}{c}}-\sqrt{\frac{c}{a}}\right)^2+\left(\sqrt{\frac{a}{c}}-\sqrt{\frac{c}{a}}\right)^2+6\ge6\)
\(\Rightarrow\)ĐPCM
Đây nè k cho mình nha:
Ta có \(\frac{a+b}{c}>\frac{a+b}{a+b+c}\)
\(\frac{b+c}{a}>\frac{b+c}{a+b+c}\)
\(\frac{a+c}{b}>\frac{a+c}{a+b+c}\)
Suy ra \(S>\frac{a+b}{a+b+c}+\frac{b+c}{a+b+c}+\frac{a+c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Vậy S > 2
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{b+c+a}+\frac{c}{c+a+b}=\frac{a+b+c}{a+b+c}=1\)
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}<\frac{a+c}{a+b+c}+\frac{b+a}{b+c+a}+\frac{c+b}{c+a+b}=\frac{2.\left(a+b+c\right)}{a+b+c}=2\)
\(1<\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}<2\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\notin N\)
\(\RightarrowĐPCM\)
Ta có: \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{c+a+b}=1\)(1)
Ta lại có \(\frac{a}{a+b}< \frac{a+c}{a+b+c}\)
=> \(a\left(a+b+c\right)< \left(a+c\right)\left(a+b\right)\)
<=> 0<bc( đúng)
CMTT: \(\frac{b}{b+c}< \frac{a+b}{a+b+c}\), \(\frac{c}{c+a}< \frac{c+b}{a+b+c}\)
Cộng lại ta được \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< 2\)(2)
Từ (1) và (2) => Tổng đó \(\notin Z\)
Vì \(\frac{a}{b}< \frac{c}{d}\)
⇒ \(ad< bc\)
⇒ \(2018ad< 2018bc\)
⇒ \(2018ad+cd< 2018bc+cd\)
⇒ \(\left(2018a+c\right)d< \left(2018b+d\right)c\)
⇒ \(\frac{2018a+c}{2018b+d}< \frac{c}{d}\)
Vậy \(\frac{2018a+c}{2018b+d}< \frac{c}{d}\) (ĐPCM)
Do \(a,b,c\in N^{\cdot}\)
\(\Rightarrow\frac{a}{a+b}>\frac{a}{a+b+c};\frac{b}{b+c}>\frac{b}{a+b+c};\frac{c}{c+a}>\frac{c}{a+b+c}\)
\(\Rightarrow1=\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\left(ĐPCM\right)\)
a) \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
b) b = a - c => b + c = a
\(\left\{{}\begin{matrix}\frac{a}{b}\cdot\frac{a}{c}=\frac{a^2}{bc}\\\frac{a}{b}+\frac{a}{c}=\frac{ac+ab}{bc}=\frac{a\left(b+c\right)}{bc}=\frac{a^2}{bc}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{c}=\frac{a}{b}+\frac{a}{c}\)
Bước 2 bạn sai rồi. Vd: \(\frac{1}{3x3}\) đâu bằng hay nhỏ hơn \(\frac{1}{2x3}\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}>\frac{a}{a+b+c}+\frac{b}{b+c+a}+\frac{c}{c+b+a}=\frac{a+b+c}{a+b+c}=1\left(1\right)\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}<\frac{2a}{b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\left(2\right)\)
\(\Rightarrow1\)<A<2=>A\(\notin N\)
=>ĐPCM