- #1
Phoenix314
- 16
- 0
a, b, c, and d are all positive real numbers.
Given that
a + b + c + d = 12
abcd = 27 + ab +ac +ad + bc + bd + cd
Determine a, b, c, and d.
---
The solution says that using AM - GM on the second equation gives
abcd (is greater than or equal to) 27 + 6*sqrt of (abcd)
From there they rewrite the second equation as:
abcd - 6sqrt(abcd) - 27 (is greater than or equal to) 0, resulting in:
sqrt(abcd) (is greater than or equal to) 9
and thus abcd ^ (1/4), which is the GM, is greater than or equal to 3
But according to AM - GM, the AM of a, b, c, and d, which is equal to 3, is greater than or equal to the GM, (abcd) ^ (1/4)
Therefore:
3 (is greater than or equal to) abcd ^ (1/4), but from previous, the GM is greater than or equal to 3. This can only occur if the AM and GM are both 3.
*The only step that I don't understand is how they applied AM - GM to
abcd = 27 + ab +ac +ad + bc + bd + cd
to obtain
abcd (is greater than or equal to) 27 + 6*sqrt of (abcd)
Any help would be appreciated! Thank you
Given that
a + b + c + d = 12
abcd = 27 + ab +ac +ad + bc + bd + cd
Determine a, b, c, and d.
---
The solution says that using AM - GM on the second equation gives
abcd (is greater than or equal to) 27 + 6*sqrt of (abcd)
From there they rewrite the second equation as:
abcd - 6sqrt(abcd) - 27 (is greater than or equal to) 0, resulting in:
sqrt(abcd) (is greater than or equal to) 9
and thus abcd ^ (1/4), which is the GM, is greater than or equal to 3
But according to AM - GM, the AM of a, b, c, and d, which is equal to 3, is greater than or equal to the GM, (abcd) ^ (1/4)
Therefore:
3 (is greater than or equal to) abcd ^ (1/4), but from previous, the GM is greater than or equal to 3. This can only occur if the AM and GM are both 3.
*The only step that I don't understand is how they applied AM - GM to
abcd = 27 + ab +ac +ad + bc + bd + cd
to obtain
abcd (is greater than or equal to) 27 + 6*sqrt of (abcd)
Any help would be appreciated! Thank you