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 4 of 5: Sum the four bounds and expand

Previous step Next step

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

Detailed analysis

Adding the four inequalities from Step 3 and factoring the left side back into $(a+2b+3c+4d)(a^2+b^2+c^2+d^2)$ , then expanding both sides as polynomials in $a,b,c,d$ , the right side equals $(a+b+c+d)^3$ minus the strictly positive term $6(abc+bcd+cda+dab)$ , so $(a+2b+3c+4d)(a^2+b^2+c^2+d^2) < (a+b+c+d)^3$ .