Hardest Query From China’s College Entrance Examination – Thoughts Your Selections

The Gaokao is China’s model of America’s SAT or ACT examination. This query from 2008 is alleged to be one of many hardest ever, as not one of the roughly 10 million test-takers solved it.

As normal, watch the video for an answer.

Hardest Query From China’s College Entrance Examination

0:00 intro
1:37 why exhausting
2:36 graph
4:01 preliminary
7:22 proof
10:47 half 1
14:20 half 2

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“All can be nicely in the event you use your thoughts to your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of sport idea and arithmetic. MindYourDecisions now has over 1,000 free articles with no advertisements due to neighborhood assist! Assist out and get early entry to posts with a pledge on Patreon.

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Reply To Hardest Query From China’s College Entrance Examination

(Just about all posts are transcribed rapidly after I make the movies for them–please let me know if there are any typos/errors and I’ll right them, thanks).

The query seems to be a calculus drawback to check discovering maxima and limiting values. However this can be very tedious to take the spinoff and second spinoff of the third time period.

I might not have solved this drawback myself. However I discovered a beautiful rationalization in this video (in Chinese language).

Substitution

Outline variables:

B = 1/√(1 + x)
C = 1/√(1 + a)

Then discover:

√[ax/(ax + 8)]
= √[1/(1 + 8/(ax))]
= 1/√(1 + 8/(ax))

So outline:

D = 1/√(1 + 8/(ax))

The issue can now be phrased within the following methodology. For a > 0, and for x > 0, we want to show:

1 < B + C + D < 2

Identities

We’ve:

B² = 1/(1 + x)

So:

1 – B² = x/(1 + x)

Since C and D have the identical type, we’ve related identities:

1 – C² = a/(1 + a)
1 – D² = (8/ax)/(1 + 8/(ax))

Thus we’ve:

(1 – B²)(1 – C²)(1 – D²) = 8B²C²D²

Since a and x are higher than 0, we are able to additionally simply discover:

0 < B < 1
0 < C < 1
0 < D < 1
B² < B
C² < C
D² < D

0 < 1 – B²
0 < 1 – C²
0 < 1 – D²

Decrease sure

For the sake of contradiction, suppose

1 ≥ B + C + D

Then we’ve:

1 – B ≥ C + D

Now let’s issue 1 – B².

1 – B²
= (1 + B)(1 – B)

Since 0 < B < 1,
1 – B²
= (1 + B)(1 – B)
> (1 – B)

Then we are able to substitute from 1 – B ≥ C + D to conclude

1 – B² > C + D

By related reasoning, we are able to derive:

1 – C² > B + D
1 – D² > B + C

As all three inequalities contain optimistic portions, we are able to multiply them and hold the instructions of the indicators. So we’ve:

(1 – B²)(1 – C²)(1 – D²)
> (C + D)(B + D)(B + C)

Now we use the AM-GM inequality that the sum of two non-negative numbers is a minimum of as giant as 2 instances the sq. root of their product.

(1 – B²)(1 – C²)(1 – D²)
> (C + D)(B + D)(B + C)
≥ 2√(CD)2√(BD)2√(BC)
= 8BCD

Recalling B, C and D are between 0 and 1, their squares can be even smaller.

8BCD
> 8B²C²D²

Thus we’ve concluded:

(1 – B²)(1 – C²)(1 – D²) > 8B²C²D²

However we had already derived this product is strictly 8B²C²D² earlier than, and we’ve reached a contradiction.

So we should have:

1 < B + C + D

Higher sure

We work equally. Suppose for the sake of contradiction that:

B + C + D ≥ 2

Then we’ve:

C + D – 1 ≥ 1 – B

As a result of C and D are between 0 and 1, we’ve:

(1 – C)(1 – D) > 0

Multiply by -1 and flip the inequality.

-(1 – C)(1 – D) < 0
-(1 – C – D + CD) < 0
-1 + C + D – CD) < 0
C + D – 1 < CD

Now let’s issue 1 – B².

1 – B²
= (1 + B)(1 – B)

Since 0 < B < 1, we’ve (1 + B) is lower than 2.

1 – B²
= (1 + B)(1 – B)
< 2(1 – B)

Now substitute C + D – 1 ≥ 1 – B after which C + D – 1 < CD.

1 – B²
= (1 + B)(1 – B)
< 2(1 – B)
< 2(C + D – 1)
< 2(CD)

So we’ve:

1 – B² < 2(CD)

We will equally derive:

1 – C² < 2(BD)
1 – D² < 2(BC)

Once more all inequalities are of optimistic portions, so we are able to multiply them and keep the course of the indicators. So we get:

(1 – B²)(1 – C²)(1 – D²)
< 2(CD)2(BD)2(BC)
= 8B²C²D²

However we had already derived this product is strictly 8B²C²D² earlier than, and we’ve reached a contradiction.

So we should have:

B + C + D < 2

References

GaoKao
https://en.wikipedia.org/wiki/Gaokao

Desmos
https://www.desmos.com/calculator/5kfgxmjrnq

N509FZ, CC BY-SA 4.0
https://w.wiki/EjLL

Kychn, CC BY-SA 4.0
https://w.wiki/EjLM

Vmenkov, CC BY-SA 3.0
https://w.wiki/EjLP

Easiest answer (in chinese language)
https://www.youtube.com/watch?v=B2EQoA-x6iM

Math StackExchange
https://math.stackexchange.com/questions/3996338/for-the-given-function-fx-frac1-sqrt-1x-frac1-sqrt-1a-sqr/3998073#3998073
https://math.stackexchange.com/questions/4165553/showing-1-frac1-sqrt1x-frac1-sqrt1a-sqrt-fracaxax82-f?noredirect=1&lq=1
https://math.stackexchange.com/questions/3354411/prove-this-inequality-via-weighted-jensen-inequality
https://math.stackexchange.com/questions/4923904/prove-inequality-1-frac1-sqrt1a-frac1-sqrt1b-frac1-sqrt1c2?noredirect=1&lq=1

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