\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)

\(\Rightarrow\frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)

\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)

\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{8.11}+...+\frac{1}{302.305}-\frac{1}{305.308}\)

\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\) 

Lời giải:

\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.205.308}\)

\(\Rightarrow \frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)

\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)

\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\)

\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)

\(\Rightarrow B< \frac{1}{40}.\frac{5}{6}\Leftrightarrow B< \frac{1}{48}\)

Bạn chép đề sai rồi, mình sửa lại đề và làm luôn nhé :

Ta có :

\(D=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)

\(\Rightarrow D=\frac{5}{6}.\left(\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\right)\)

\(\Rightarrow D=\frac{5}{6}.\left(\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\right)\)

\(\Rightarrow D=\frac{5}{6}.\left(\frac{1}{5.8}-\frac{1}{305.308}\right)\)

\(\Rightarrow D=\frac{5}{6}.\frac{1}{40}-\frac{5}{6}.\frac{1}{305.308}\)

\(\Rightarrow D=\frac{1}{48}-\frac{5}{6.305.308}< \frac{1}{48}\) (đpcm )

Áp dụng BĐT \(AM-GM\) ta có :

\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2\ge5\sqrt[5]{\dfrac{a^{15}b^4}{b^9}}=5\dfrac{a^3}{b}\)

\(\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+c^2+c^2\ge5\sqrt[5]{\dfrac{b^{15}c^4}{c^9}}=5\dfrac{b^3}{c}\)

\(\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+a^2+a^2\ge5\sqrt[5]{\dfrac{c^{15}a^4}{a^9}}=5\dfrac{c^3}{a}\)

Cộng từng vế của BĐT ta được :

\(3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(a^2+b^2+c^2\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)

Tiếp tục áp dụng BĐT \(AM-GM\) ta lại có :

\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2+b^2\ge5\sqrt[5]{\dfrac{a^{10}b^6}{b^6}}=5a^2\)

\(\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+c^2+c^2+c^2\ge5\sqrt[5]{\dfrac{b^{10}c^6}{c^6}}=5b^2\)

\(\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+a^2+a^2+a^2\ge5\sqrt[5]{\dfrac{c^{10}a^6}{a^6}}=5c^2\)

Cộng vế theo vế ta được :

\(2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+3\left(a^2+b^2+c^2\right)\ge5\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge2\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)

\(\Rightarrow3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(a^2+b^2+c^2\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)

\(\Leftrightarrow5\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)

\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\left(đpcm\right)\)