Problem 2

Let

a

b

c

d

be real numbers such that

a \ge b \ge c \ge d > 0

and

a+b+c+d=1

. Prove that

(a+2b+3c+4d)\,a^ab^bc^cd^d < 1

Step 3 of 5: Majorize each term

a^2,b^2,c^2,d^2

separately using the ordering

a\ge b\ge c\ge d

Previous step Next step

In plain words

Since $a,b,c,d$ are decreasing, the coefficient $4$ on $d$ (the smallest variable) can be replaced by a $3$ shifted onto a larger variable without decreasing the sum, giving four separate upper bounds for $a+2b+3c+4d$ , one for use with each of $a^2,b^2,c^2,d^2$ .

a^2(a+2b+3c+4d)\le a^2(a+3b+3c+3d),\ \ b^2(a+2b+3c+4d)\le b^2(3a+b+3c+3d),\ \ c^2(a+2b+3c+4d)\le c^2(3a+3b+c+3d),\ \ d^2(a+2b+3c+4d)\le d^2(3a+3b+3c+d)

Detailed analysis

Since $d\le b$ , $a+2b+3c+4d\le a+3b+3c+3d$ . Since $b+d\le 2a$ (from $a\ge b$ and $a\ge d$ ), $a+2b+3c+4d\le 3a+b+3c+3d$ . Since $2c+d\le 2a+b$ (from $a\ge c$ twice and $b\ge d$ ), $a+2b+3c+4d\le 3a+3b+c+3d$ . Since $2a+b\ge 3d$ (from $a\ge d$ twice and $b\ge d$ ), $a+2b+3c+4d\le 3a+3b+3c+d$ . Multiplying each by the corresponding nonnegative $a^2,b^2,c^2,d^2$ gives the four stated inequalities.