问题设m+n=3,求m·3m+n·3n的最小值.这是《数学通讯》2013年第11、12期P65张乃贵老师提到的一道条件最值问题.由于条件和目标函数都成对称性,容易猜测最值为9槡3,但不易证明.通过探究,下面给出两种新颖简洁的解法,供参考.解法1不妨设m≥n,则m·3m+n·3n-(m+n)·3m+n/2=3m+n/2·(m·3m-n/2+n·3n-m/2-m-n)=3m+n/2·[m(3m-n/2-1)+n(3n-m/2-1)] The problem is to set m + n = 3, and find the minimum value of m · 3m + n · 3n, which is the most significant condition mentioned by Teacher Zhang Naigui in “Mathematical Communication”, Items 11 and 12, 2013. Due to the condition and objective function Are symmetrical, easy to guess the value of 9 槡 3, but not easy to prove.Through exploration, two new and concise solutions are given below for reference.Method 1 Let m ≥ n, then m · 3m + n · 3n - (m + n) · 3m + n / 2 = 3m + n / 2 · (m · 3m-n / 2 + n · 3n-m / 2-mn) = 3m + n / 2 · [m n / 2-1) + n (3n-m / 2-1)]