Giả sử a< b < c thì a \(\ge\)2 , b \(\ge\)3 , c\(\ge\)5 . Ta có :

\(\frac{1}{\left[a,b\right]}=\frac{1}{ab}\le\frac{1}{6},\frac{1}{\left[c,a\right]}=\frac{1}{ca}\le\frac{1}{10}\)

=> vế trái nhỏ hơn hoặc bằng \(\frac{1}{6}+\frac{1}{15}+\frac{1}{10}=\frac{1}{3}\)

Bài 1a:

A = \(\frac{7^2}{7.8}\times\frac{8^2}{8.9}\times\ldots\times\frac{11^2}{11.12}\)

A = \(\frac{7.8.9.\ldots11}{7.8.\ldots11}\) x \(\frac{7.8.9\ldots11}{8.9.12}\)

A = 7/12

Bài 1B

B = (1+ 1/11).(1 + 1/12)...(1+ 1/15)

B = (11/11 + 1/11).(12/12 + 1/12)...(15/15+ 1/15)

B = 12/11.13/12....16/15

B = 16/11

\(B=\frac{12}{11}x\frac{13}{12}x.......x\frac{16}{15}\)

\(=\frac{16}{11}\)

Bài 1a:

A = \(\frac{7^2}{7.8}\times\frac{8^2}{8.9}\times\ldots\times\frac{11^2}{11.12}\)

A = \(\frac{7.8.9.\ldots11}{7.8.\ldots11}\) x \(\frac{7.8.9\ldots11}{8.9.12}\)

A = 7/12

Giả sử \(a< b< c\)thì \(a\ge2\)\(;\)\(b\ge3\)\(;\)\(c\ge5\)

Ta có:

\(\frac{1}{\left[a,b\right]}=\frac{1}{ab}\le\frac{1}{6}\)\(;\)\(\frac{1}{\left[b,c\right]}=\frac{1}{bc}\le\frac{1}{15}\)\(;\)\(\frac{1}{\left[c,a\right]}=\frac{1}{ca}\le\frac{1}{10}\)

Do đó: \(\frac{1}{\left[a,b\right]}+\frac{1}{\left[b,c\right]}+\frac{1}{\left[c,a\right]}\le\)\(\frac{1}{6}+\frac{1}{15}+\frac{1}{10}=\frac{1}{3}\)

\(\Rightarrow\)\(\frac{1}{\left[a,b\right]}+\frac{1}{\left[b,c\right]}+\frac{1}{\left[c,a\right]}\le\)\(\frac{1}{3}\)\(\rightarrowĐPCM\)

\(\text{Vì }\left[a,b\right],\left[b,c\right],\left[c,a\right]\text{ là BCNN}\)

\(\Rightarrow\left[a,b\right]=a.b;\left[b,c\right]=b.c;\left[c,a\right]=c.a\)

\(\Rightarrow\frac{1}{\left[a+b\right]}+\frac{1}{\left[b+c\right]}+\frac{1}{\left[c+a\right]}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

\(\text{Giả sử }a< b< c\)

\(\Rightarrow a\le2;b\le3;c\le5\)

\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\le\frac{1}{2.3}+\frac{1}{3.5}+\frac{1}{5.2}=\frac{1}{3}\)

\(\text{hay }\frac{1}{\left[a+b\right]}+\frac{1}{\left[b+c\right]}+\frac{1}{c+a}\le\frac{1}{3}\left(đpcm\right)\)

ể ==

\(2< 3\Rightarrow\frac{1}{2}>\frac{1}{3}\)

Cậu Bé Tiến Pro: e đổi dấu đi :))