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 5 of 5: Conclude the inequality using

a+b+c+d=1

Previous step Next step

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

Detailed analysis

Combining Step 2 and Step 4 with the hypothesis $a+b+c+d=1$ gives $(a+2b+3c+4d)a^ab^bc^cd^d\le(a+2b+3c+4d)(a^2+b^2+c^2+d^2)<(a+b+c+d)^3=1$ , which is exactly the required inequality. Equality in the first step forces $a=b=c=d=\frac14$ , at which point the chain gives the strict bound $\frac58<1$ , confirming the inequality is strict.