找一组满足条件的正数怎么这么难呢？

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已知$a，b，c，d＞0，a+b+c+d=2，\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}=4,$

请找出一组满足条件的a、b、c、d的值，或者给出a、b、c、d的参数表达式。

hbghlyj

$A = a + b + c, B = b + c + d, C = c + d + a, D = d + a + b$.$$(A + B + C + D)\left(\frac{1}{A} + \frac{1}{B} + \frac{1}{C} + \frac{1}{D}\right) \geq (1 + 1 + 1 + 1)^2 = 16.$$
$$(a + b + c + d)\left(\frac{1}{a + b + c} + \frac{1}{b + c + d} + \frac{1}{c + d + a} + \frac{1}{d + a + b}\right)\ge\frac{16}3$$

hbghlyj

$a = b=c=\frac{1}{4+\sqrt{10}},d = \frac{\sqrt{10}}2$
sagecell.sagemath.org?q=bkdhih

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hbghlyj 发表于 2025-5-29 22:40

$a = b=c=\frac{1}{4+\sqrt{10}},d = \frac{\sqrt{10}}2$

sagecell.sagemath.org?q=bkdhih

a、b、c、d的值，刚才是对的，现在d的值不对。

其实都对：
$a = b=c=\frac{1}{4+\sqrt{10}},d = \frac{5+2\sqrt{10}}{4+\sqrt{10}}=\frac{\sqrt{10}}2$

经验证，符合题设的两个条件。

hbghlyj

走走看看 发表于 2025-5-29 15:43
a、b、c、d的值，刚才是对的，现在d的值不对。

$a = b = c = \frac{1}{4+\sqrt{10}},d = \frac{\sqrt{10}}{2}$

$a+b+c+d = \frac{3}{4+\sqrt{10}} + \frac{\sqrt{10}}{2}$

$\frac{3}{4+\sqrt{10}} = \frac{3(4-\sqrt{10})}{(4+\sqrt{10})(4-\sqrt{10})}$
$(4+\sqrt{10})(4-\sqrt{10}) = 4^2 - (\sqrt{10})^2 = 16 - 10 = 6$

∴$\frac{3}{4+\sqrt{10}} = \frac{3(4-\sqrt{10})}{6} = \frac{4-\sqrt{10}}{2}$

$a+b+c+d = \frac{4-\sqrt{10}}{2} + \frac{\sqrt{10}}{2}= 2$

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$a+b+c = \frac{1}{4+\sqrt{10}} + \frac{1}{4+\sqrt{10}} + \frac{1}{4+\sqrt{10}} = \frac{3}{4+\sqrt{10}}$

$b+c+d = \frac{1}{4+\sqrt{10}} + \frac{1}{4+\sqrt{10}} + \frac{\sqrt{10}}{2} = \frac{2}{4+\sqrt{10}} + \frac{\sqrt{10}}{2}$

$\frac{2}{4+\sqrt{10}} = \frac{2(4-\sqrt{10})}{(4+\sqrt{10})(4-\sqrt{10})} = \frac{2(4-\sqrt{10})}{6}$

$b+c+d = \frac{2(4-\sqrt{10})}{6} + \frac{3\sqrt{10}}{6} = \frac{8-2\sqrt{10}+3\sqrt{10}}{6} = \frac{8+\sqrt{10}}{6}$

∴$\frac{1}{b+c+d} = \frac{1}{\frac{8+\sqrt{10}}{6}} = \frac{6}{8+\sqrt{10}} = \frac{6(8-\sqrt{10})}{(8+\sqrt{10})(8-\sqrt{10})} = \frac{6(8-\sqrt{10})}{8^2 - (\sqrt{10})^2} = \frac{6(8-\sqrt{10})}{64-10} = \frac{6(8-\sqrt{10})}{54} = \frac{8-\sqrt{10}}{9}$

$\frac{1}{b+c+d} = \frac{1}{c+d+a} = \frac{1}{d+a+b} = \frac{8-\sqrt{10}}{9}$

$\frac{1}{a+b+c} + \frac{1}{b+c+d} + \frac{1}{c+d+a} + \frac{1}{d+a+b}$
$= \frac{4+\sqrt{10}}{3} + 3 \left(\frac{8-\sqrt{10}}{9}\right)$
$= \frac{4+\sqrt{10}}{3} + \frac{8-\sqrt{10}}{3}$
$= \frac{4+\sqrt{10} + 8-\sqrt{10}}{3}$
$= \frac{4+8}{3}$
$= \frac{12}{3}$
$= 4$