Vì \(0< a1< a2< a3< ...< a15\)nên ta có:
\(\hept{\begin{cases}a1+a2+a3+a4+a5< 5a5\\a6+a7+a8+a9+a10< 5a10\\a11+a12+a13+a14+a15< 5a15\end{cases}\Rightarrow\frac{a1+a2+a3+...+a15}{a5+a10+a15}< \frac{5.\left(a5+a10+a15\right)}{a5+a10+a15}=5}\)
Vậy...
Ta có:a1<a2<a3<......,a15 =>a1+a2+...+a5<5a5;
a6+a7+...........+a10<5a10
a11+a12+.....+a15<5a15
=>a1+a2+a3+....+a15<5(a5+a10+a15)
=\(\frac{a1+a2+a3+....+a15}{a5+a10+a15}\)<5
Ta có \(a_1+a_2+a_3+a_4+a_5< 5\)
\(a_6+a_7+a_8+a_9+a_{10}< 5a_{10}\)
\(a_{11}+a_{12}+a_{13}+a_{14}+a_{15}< 5a_{15}\)
\(\Rightarrow a_1+a_2+....+a_{15}< 5\left(a_5+a_{10}+a_{15}\right)\)
Vậy \(\frac{a_1+a_2+a_3+...+a_{15}}{a_5+a_{10}+a_{15}}< 5\)
Đúng ko các bạn? có ai có cách làm khác k ?
ta có
a1+a2+a3+a4+a5>5a5
a6+a7+...+a10>5a10
a11+..+a15>5a15
=>a1+a2+...+a15>5(a5+a10+a15)
-=>đpcm
Ta có:Vi\(0< a_1< a_2< a_3< ...< a_{15}\)
Nên \(\frac{a_1+a_2+a_3+...+a_{15}}{a_5+a_{10}+a_{15}_{ }}=\frac{\left(a_1+a_2+a_3+a_4+a_5\right)+\left(a_6+a_7+a_8+a_9+a_{10}\right)+\left(a_{11}+a_{12}+a_{13}+a_{14}+a_{15}\right)}{a_5+a_{10}+a_{15}}\)
\(< \frac{\left(a_5+a_5+a_5+a_5+a_5\right)+\left(a_{10}+a_{10}+a_{10}+a_{10}+a_{10}\right)+\left(a_{15}+a_{15}+a_{15}+a_{15}+a_{15}\right)}{a_5+a_{10}+a_{15}}\)
\(=\frac{5\times a_5+5\times a_{10}+5\times a_{15}}{a_5+a_{10}+a_{15}}=\frac{5\times\left(a_5+a_{10}+a_{15}\right)}{a_5+a_{10}+a_{15}}=5\)
Vậy ........................
Chúc bạn học tốt
Ta có:\(a_1< a_2< a_3< a_4< a_5\Rightarrow a_1+a_2+a_3+a_4+a_5< 5\cdot a_5\left(1\right)\)
Tương tự:\(a_6+a_7+a_8+a_9+a_{10}< 5a_{10}\left(2\right)\)
\(a_{11}+a_{12}+a_{13}+a_{14}+a_{15}< 5\cdot a_{15}\left(3\right)\)
Từ (1) (2) (3) => đặt chung=> chuyển vế=> đpcm
