Tough Olympiad-like Inequalities question

[Phoenix314]

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.

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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

 

[VietDao29]

*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

Hmm, you should note that:
$$ab + cd \geq 2 \sqrt{abcd}$$
$$ac + bd \geq 2 \sqrt{abcd}$$
$$ad + bc \geq 2 \sqrt{abcd}$$
Adding both sides of the 4 inequalities, we have:
$$ab +ac +ad + bc + bd + cd \geq 6 \sqrt{abcd}$$
So that means:
$$abcd = 27 + ab +ac +ad + bc + bd + cd \geq 27 + 6\sqrt{abcd}$$
Can you get this? :)

 

[0rthodontist]

I got it in 1 minute by guessing.