như câu trả lời của bạn Phuc Tran

vì \(\frac{1}{4}< 1,\frac{1}{5}< 1,......,\frac{1}{19}< 1\)  nên B < 1.

Ta có: \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)

\(\Rightarrow B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)

Vì \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=5\cdot\frac{1}{9}=\frac{5}{9}>\frac{1}{2}\)

Vì \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+...+\frac{1}{19}=10\cdot\frac{1}{19}=\frac{10}{19}>\frac{1}{2}\)

\(\Rightarrow B>\frac{1}{4}+\frac{5}{9}+\frac{10}{19}>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}=\frac{1}{4}+\frac{2}{4}+\frac{2}{4}\)

\(\Rightarrow B>\frac{5}{4}>1\Rightarrow B>1\)

Ta có : B = 1/3 + 1/3^2 + 1/3^3 +...+ 1/3^99

=> 3B - B = ( 1 + 1/3 + 1/3^2 +...+ 1.3^99) - ( 1/3 + 1/3^2 + 1/3^3 +...+ 1/3^99 )

=> 2B = 1 - 1/3^99 < 1

=> 2B < 1

=> B < 1/2 ( ĐPCM )

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)

\(B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)

\(B< 1-\frac{1}{8}< 1\left(dpcm\right)\)

Đặt a-1=x, b-1=y (\(x,y>\frac{\sqrt{5}-3}{2}\))

=> \(xy=1\)

VT= \(\frac{1}{\left(x+1\right)^2+x}+\frac{1}{\left(y+1\right)^2+y}=\frac{1}{\left(\frac{1}{y}+1\right)^2+\frac{1}{y}}+\frac{1}{\left(y+1\right)^2+y}=\frac{y^2+1}{\left(y+1\right)^2+y}\)\(=\frac{2}{5}-\frac{3\left(y-1\right)^2}{\left(y+1\right)^2+y}\ge\frac{2}{5}\)(do \(\left(y+1\right)^2+y=b^2+b-1>0\))

Dấu bằng khi \(x=y=1\)=> \(a=b=2\)

đơn giản hơn cách của quý đây

a+b=ab => \(\frac{1}{a}+\frac{1}{b}=1\)Đặt \(\frac{1}{a}=x;\frac{1}{y}=b\)

Khi đó \(\frac{1}{a^2+a-1}=\frac{1}{\left(\frac{1}{x}\right)^2+\frac{1}{x}-1}=\frac{x^2}{1+x-x^2}\)

Chứng minh tương tự với b

=> Đặt A=\(\frac{1}{a^2+a-1}+\frac{1}{b^2+b-1}=\frac{x^2}{1+x-x^2}+\frac{y^2}{1+y-y^2}\)

Cauchy-Schwarz và nhớ: x+y=1 và x²+y² >=1/2

OK

như câu trả lời của bạn Phuc Tran

Ta có :

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+..............+\frac{1}{19}\)

\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+.........+\frac{1}{19}\right)\)

Ta thấy :

\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{1}{9}.5=\frac{5}{9}>\frac{1}{2}\)

\(\frac{1}{10}+\frac{1}{11}+....+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{1}{19}.5>\frac{10}{19}>\frac{1}{2}\)

\(\Rightarrow B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}>1\)

ta có

1/4+1/5+....+1/10>1/10,7=7/10

1/11+1/12+.....+1/19>1/20,9=9/20

kết hợp lại ta có b=1/4+1/5+1/6+......+1/19>7/10+9/20=23/20>1

vậy b>1