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The problem is very weird, first said a, b, c, d are all greater than 0, but then there is -1 4 should mean that they are all real numbers.
If they are all real numbers, there is.
abcd|=|ab|*|cd| <= [(ab)^2+(cd)^2}/2 <= [(a^2+b^2)^2/4 + c^2+d^2)^2/4]/2=1/4
i.e. |abcd|<=1/4
1/4<=abcd<=1/4
When|a|=|b|=|c|=|d|= Half of the root number, two times take the equal sign.
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Upstairs, you're so good, I dizzy after reading it, I really won't, some questions belong to you.
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Noting that the values a, b, and c are required for the radical simple roll, then the conjecture is the case where the equal sign is taken. So the idea is: prove the left side and the right side, and then take the etc. The most basic method is recipes.
First of all, notice that there is an absolute value, and put the non-absolute value to the absolute value, i.e
12|a||b|+4|b||c|-16c-16 Mingyu 12a|b|-4b|c|-16c-16
Also note (12a +7b +5c )-12|a||b|+4|b||c|-16c-16)=(12a²-12|a||b|+3b²)+4b²-4|b||c|+c²)+4c²+16c+16)……Note: This step needs to be observed by yourself, and to put it bluntly, it is made up.
Continued from the above equation:)=3(2|a|-|b|)²2|b|-|c|)²4(c+2)²≥0
12a²+7b²+5c²≥12|a||b|+4|b||c|-16c-16
And 12|a||b|+4|b||c|-16c-16≥12a|b|-4b|c|-16c-16
12a²+7b²+5c²≥12a|b|-4b|c|-16c-16
From the inscription, 12a +7b +5c 12a|b|-4b|c|-16c-16
12a +7b +5c =12a|b|-4b|c|-16c-16
Investigate the conditions for each inequality deflation (i.e., 3(2|a|-|b|)²2|b|-|c|The inequality of 4(c+2) 0 takes the equality condition, and 12|a||b|+4|b||c|-16c-16≥12a|b|-4b|c|-16c-16).
The former requires 2|a|=|b|,2|b|=|c|, c = -2, the latter requires a 0, b 0
Therefore, a=1 2, b=-1, c=-2 can be obtained
After the solution, you can substitute the test if you don't worry.
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│a│-2│b│+½c│
Hope to give you a correct answer!
Good luck with your studies!
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Shift the fuel to do the item, get:
1 2) (a b )-Pi Sanheng a+b) 2] Digging dry.
2(a²+b²)]1/4)[(a+b)²
1/4)[a²-2ab+b²]
1/4)(a-b)²≥0
i.e.: 1 2) (a b ) a+b) 2].
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Are you sure you didn't make a mistake? It should be a= 12, b=-7, right?
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