Question

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Cho 6 số nguyên dương a < b < c < d < m <n. Chứng minh: \(\frac{a+d}{a+b+c+m+n}< \frac{1}{3}\)

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Answer 1

[Khánh Ngọc]

a < b < c < d < m

=> a + d < c + m + n

=> 3 ( a + d ) < a + b + c + d + m + n

\(\Rightarrow\frac{3\left(a+d\right)}{a+b+c+d+m+n}< 1\)

\(\Rightarrow\frac{a+d}{a+b+c+d+m+n}< \frac{1}{3}\) ( Đpcm )