MathLabs

Problem 2

Let aa, bb, cc, dd be real numbers such that a≥b≥c≥d>0a \ge b \ge c \ge d > 0 and a+b+c+d=1a+b+c+d=1. Prove that (a+2b+3c+4d) aabbccdd<1(a+2b+3c+4d)\,a^ab^bc^cd^d < 1.
Step 2 of 5: Reduce to bounding (a+2b+3c+4d)(a2+b2+c2+d2)(a+2b+3c+4d)(a^2+b^2+c^2+d^2)
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(a+2b+3c+4d) aabbccdd≤(a+2b+3c+4d)(a2+b2+c2+d2)(a+2b+3c+4d)\,a^ab^bc^cd^d\le (a+2b+3c+4d)(a^2+b^2+c^2+d^2)
Detailed analysis

Multiplying Step 1's inequality by the positive quantity a+2b+3c+4da+2b+3c+4d gives (a+2b+3c+4d)aabbccdd≤(a+2b+3c+4d)(a2+b2+c2+d2)(a+2b+3c+4d)a^ab^bc^cd^d\le (a+2b+3c+4d)(a^2+b^2+c^2+d^2), so it suffices to show the right side is strictly less than 11.