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 1 of 5: Bound

a^ab^bc^cd^d

by weighted AM–GM

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In plain words

Treating $a,b,c,d$ as both the weights and the values in weighted AM–GM directly compares the weighted power mean $a^ab^bc^cd^d$ to the ordinary quadratic mean $a^2+b^2+c^2+d^2$ , since $a+b+c+d=1$ .

\frac{a\cdot a+b\cdot b+c\cdot c+d\cdot d}{a+b+c+d}\ge \sqrt[a+b+c+d]{a^ab^bc^cd^d}\ \implies\ a^ab^bc^cd^d\le a^2+b^2+c^2+d^2

Detailed analysis

By weighted AM–GM with weights $a,b,c,d$ (summing to $1$ ) applied to the values $a,b,c,d$ themselves, $\frac{a\cdot a+b\cdot b+c\cdot c+d\cdot d}{a+b+c+d}\ge a^ab^bc^cd^d$ , i.e. $a^ab^bc^cd^d\le a^2+b^2+c^2+d^2$ .