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Question

chứng minh rằng với

$a, b, c$ dương:

$\frac{1}{a+2b+c} +\frac{1}{b+2c+a}+\frac{1}{c+2a+b}\leq \frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}$

score 4 · Answer 1

Bổ đề:
Với

$a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n>0$ , ta có:

$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\ldots+\frac{a_n^2}{b_n}\ge\frac{(a_1+a_2+\ldots+a_n)^2}{b_1+b_2+\ldots+b_n}$

Áp dụng ta có:

$\frac{4}{a+3b}+\frac{2}{b+3c}+\frac{1}{c+3a}=\frac{16}{4(a+3b)}+\frac{4}{2(b+3c)}+\frac{1}{c+3a}$

$\ge\frac{49}{7a+14b+7c}=\frac{7}{a+2b+c}$
Tương tự:

$\frac{1}{a+3b}+\frac{4}{b+3c}+\frac{2}{c+3a}\ge\frac{7}{a+b+2c}$

$\frac{2}{a+3b}+\frac{1}{b+3c}+\frac{4}{c+3a}\ge\frac{7}{2a+b+c}$
Cộng các BĐT trên lại ta được đpcm.

1 Đáp án