Problem:
Find the minimum possible value of
b3+4a​+c3+4b​+d3+4c​+a3+4d​
given that a,b,c,d are nonnegative real numbers such that a+b+c+d=4.
Solution:
(Proposed by Titu Andreescu)
We will show that the minimum is 32​. (In particular, the value 54​, obtained by making the natural guess a=b=c=d=1, is not the right answer.)
We have
b3+44a​=a−b3+4ab3​≥a−3ab​
since
b3+4=2b3​+2b3​+4≥3b2
by the Arithmetic Mean-Geometric Mean Inequality.
Then
b3+4a​+c3+4b​+d3+4c​+a3+4d​≥4a+b+c+d​−12ab+bc+cd+da​
But a+b+c+d=4 and
4(ab+bc+cd+da)=4(a+c)(b+d)≤(a+b+c+d)2=16
Hence
b3+4a​+c3+4b​+d3+4c​+a3+4d​≥1−124​=32​
The minimum is realized when, for example, a=b=2 and c=d=0.
The problems on this page are the property of the MAA's American Mathematics Competitions