[Guest]

I was puzzled by this one..

Show that for positive a, b, c, and d, such that abcd=1, a^2 +b^2+c^2+d^2 + ab+ac+ad+bc+bd+cd is not smaller then 10.

[Guest]

a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd

=1/3[(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(d-c)^2]+5/3[{(ab)^0.5-(cd)^0.5}^2+ {(ac)^0.5-(bd)^0.5}^2 +{(ad)^0.5-(bc)^0.5}^2] + 10(abcd)^0.5

=Positive Number or Zero + 10 (abcd)^0.5 (square of a no. is greater than or equal to zero, (x-y)^2 >=0)

=Positive Number or Zero + 10 (abcd=1)

Hence Proved!!!

[Guest]

ashish_soni: I'm confused how you got your first step. Those equations don't appear to be equal... I could be missing something...

My calculations:

[unreality]

I was puzzled by this one..

Show that for positive a, b, c, and d, such that abcd=1, a^2 +b^2+c^2+d^2 + ab+ac+ad+bc+bd+cd is not smaller then 10.

abcd = 1

F(a,b,c,d) = (aa + bb + cc + dd + ab + ac + ad + bc + bd + cd) >= 10

I think the only way to solve this is to make it a calculus minimization problem where we find the smallest possible value of F satisfying the abcd=1 condition, and if that value is 10 or higher, the postulation has been proven.

So first we need to find the derivative of F. I don't know how to do derivatives of functions with multiple input so I'm just going to do it differential-style on all the variables and see where that gets me haha. I'm gonna use 'd' for diferential though and switch the d variable to e to make it easier to read though

F'(a,b,c,e) * dF = 2ada + 2bdb + 2cdc + 2ede + adb + bda + ade + eda + bdc + cdb + bde + edb + cde + edc

dividing all by dF and setting equal to zero, and using notation of a' to mean da/dF, etc.

0 = 2a a' + 2b b' + 2c c' + 2e e' + a b' + b a' + a e' + e a' + b c' + c b' + b e' + e b' + c e' + e c' = F'(a,b,c,e)

also knowing that abce=1, we have to find a set of values that make F'(a,b,c,e) equal to zero. For all such sets then we must find the minimum value of F from those.

the problem can be simplified into finding values that satisfy this:

0 = 2a + 2b + 2c + 2e + a + b + a + e + b + c + b + e + c + e

0 = 4a + 4b + 4c + 4e

0 = a + b + c + e

we know that 1 = abce

so I guess now it's just intuition. If we have { 1, 1, -1, -1 } that would work. There might be others but let's plug that into F and start with that.

ab + ac + ad + bc + bd + cd

F(1,1,-1,-1) = 1 + 1 + 1 + 1 + 1 - 1 - 1 - 1 -1 + 1 = 2

It seems that we've disproven it.

except damn, didn't see that they all ahd to be POSITIVE.

Okay back to the drawing board:

a + b + c + d = 0

abcd = 1

we have to find the solution sets to this and plug them back into F and always get something 10 or higher.

except the problem is that I clearly messed up somewhere because if they're all positive then a+b+c+d = 0 is impossible.

I'll come back to this maybe haha

[Guest]

using AM>=GM

AM is arithematis mean and GM is geometric mean.

(a+b+c+d)/4>=(abcd)^1/4

as abcd=1 and taking squares on both sides...

aa+bb+cc+dd+2ab+2ac+2ad+2bc+2bd+2cd>=16

divide full inequality by 2 we get

aa/2+bb/2+cc/2+dd/2+ab+ac+ad+bc+bd+cd>=8

minimum value of aa+bb+cc+dd will be 4 when all a,b,c and d will be 1 as any other value will increase the value as abcd=1 and if we take a=.5 then anyone will be 2 which will increase the value.

so by small arrangmetns we get the answer

[Guest]

littlej:It is not {(ad)^0.5+(bc)^0.5}^2, but {(ad)^0.5-(bc)^0.5}^2. You made a mistake there.

If you expand my equation(all the squares), you will get back the original equation(g).

[Guest]

one more thing in the last step of solving it is 10+positive number, since a.b.c.d=1 (give in the puzzle itself)