Fun with real numbers

[BMAD]

I. suppose that for all reals 0 <= a <= b <= c <= d , we have: (a + b + c + d)^2 >= K b c . Find the largest possible value of K

II. Suppose a + b + c = 1 , where a , b , and c are non-neg. real nos. Prove that we always have: 7(a b + a c + b c) <= 2 + 9 a b c .

[plasmid]

For part 1: If you have (a+b+c+d)² >= Kbc for all such reals,

Then you can just consider the "worst case" where a=0 and d=c. If K is small enough that the inequality is true for those values of a and d, then any other values of a and d would just make the left side greater so the inequality would still be true.

 

So considering the worst case:

(b + 2c)² >= Kbc

b² + 4bc + 4c² >= Kbc

b/c + 4 + 4c/b >= K

 

Define w as b/c, and you have

w + 4 + 4/w >= K

 

To find the minimum value of the left hand side, solve the derivative d/dw and set it to zero

d/dw = 1 – 4/(w²)

d/dw = 0 when w = 2

 

Going back to the inequality that we just took the derivative of

w + 4 + 4/w >= K

2 + 4 + 2 >= K

8 >= K

 

So the largest possible value of K is 8.

Edited July 28, 2015 by plasmid

[araver]

For part I. 

I get K=9.

The inequality holds for K=0.

Assume there is a largest K for which the inequality holds for all 0 <= a <= b <= c <= d. Then for some a=0, b=c=d > a= 0 we get (a + b + c + d)^2 >= K b c <=> (3d)^2 >=Kd^2. Since d>0, this means K<=9.

For all 0 <= a <= b <= c <= d the following holds true:

(a + b + c + d)^2 >= (0 + b + c + c)^2 = (b+2*c)^2 = b^2 + (2*c)^2 + 2*b*(2*c) = b^2 + c^2 + 3*c^2 + 2*b*c + 2*b*c =(b+c)^2 + 3*c^2 +2*b*c >= 4*b*c + 3*b*c + 2*b*c = 9*b*c.

Last step uses at once:

- (b+c)^2 >= 4bc which is equivalent to (b-c)^2 >=0

- 3*c^2 = 3*c * c >= 3*b*c since c>=b, c,b>=0

Equality holds for a=0, d=c, b=c.

 

 

 

[plasmid]

 Araver's right.

 

When I solved for the derivative of w and found that d/dw = 0 at w = 2, I didn't take into account that w can't be 2 in the first place since I defined w as b/c, and b <= c. So I should have just considered the extreme values of w at w=0 and w=1.

w=0 makes the 4/w term go to infinity so it's obviously not the lower bound.

w=1 gives w + 4 + 4/w >= K, so 9 >= K, which is what araver got.

Edited July 29, 2015 by plasmid

[araver]

Answer for Part II

 We can easily see if a=b=c=1/3 the equality holds. So we're looking for inequalities that hold for equal numbers.
 
 Thinking of where abc appears as a term, a few possibilities pop into mind:
 1) the AM-GM inequality: (a+b+c)^3 >= 27abc. Or a^3 + b^3 + c^3 >= 3 (a^3 b^3 c^3)^(1/3) = 3abc.
 2) we use 1 = (a+b+c)^3 which contains abc.
 
 First approach did not get me anywhere since abc is the lesser term of the inequalities.

 
 Using the second, the hard way i.e. rewriting the right term (R):
 2+9abc = 2(a+b+c)^3 + 9abc = 2(a+b+c)(a^2+b^2+c^2+2ab+2ac+2bc) + 9abc = 
 2(a^3+b^3+c^3+3ba^2+3ab^2+3ca^2+3ac^2+3bc^2+3cb^2+6abc) + 9abc = 
 2(a^3+b^3+c^3)+6(ba^2+ab^2+ca^2+ac^2+bc^2+cb^2)+21abc =
 2(a^3+b^3+c^3) + 6 [ ab(a+b+c) + bc (a+b+c) + ca (a+b+c)] + 3abc =
 2(a^3+b^3+c^3) + 6 (ab+bc+ca)(a+b+c) + 3abc =
 6 (ab+bc+ca) +  2(a^3+b^3+c^3) + 3abc = R
 
 Now we have something very similar to the left term but with a factor of 6 instead of 7.

Denote L =  7(ab + ac + bc) and using the same trick to get from (ab + ac + bc) to create and simplify abc:
 R - L = 2(a^3+b^3+c^3) + 3abc - (ab + ac + bc) = 
 2(a^3+b^3+c^3) + 3abc - (ab + ac + bc) (a+b+c) =
 2(a^3+b^3+c^3) + 3abc - (ba^2+ab^2+ca^2+ac^2+bc^2+cb^2 + 3abc) =
 2(a^3+b^3+c^3) - (ba^2+ab^2+ca^2+ac^2+bc^2+cb^2) = T

Got rid of abc and this looks symmetrical enough to try a brute approach e.g. find an inequality where ba^2 is on the smaller term of the inequality.

AM-GM comes into mind again:
ba^2 = baa = (b^3 a^3 a^3)^(1/3) <= (b^3 + a^3 + a^3)/3  = b^3/3 + 2 a^3/3 with equality iff a=b
Summing this inequality with the other 5 symmetrical AM-GM inequalities we get T >=0 (since each x^3 appears twice as 2/3 and twice as 1/3)

So L<=R with equality iff a=b=c=1/3

q.e.d.