Can't we just do greedy algorithm for (the minimal) sum of the elements?

There is a simpler way. Since $(a+b)^2 \ge 4ab$ for all $a,b\in \mathbb R$, the product of the numbers on the board is multiplied by at least $4$ at each move, so $m^n \ge 4^h$, which rearranges to the desired.

http://artofproblemsolving.com/community/c6h1113184p5083546

rafayaashary1 wrote: Can't we just do greedy algorithm for (the minimal) sum of the elements? Can you please elaborate your solution? Thanks.

When we operate on two numbers, the overall sum increases by the sum of those two numbers. So then to keep it minimal, we can operate on the two minimal elements, and the overall set of numbers will consist of $2^k$, $2^{k+1}$, and $2^k+1$ for some integer $k$ where that last part comes from when $k$ is odd.

rafayaashary1 wrote: When we operate on two numbers, the overall sum increases by the sum of those two numbers. So then to keep it minimal, we can operate on the two minimal elements, and the overall set of numbers will consist of $2^k$, $2^{k+1}$, and $2^k+1$ for some integer $k$ where that last part comes from when $k$ is odd. But it required each number to be equal to $m$ at the end. What do you do next do prove $h \le \tfrac 12 n \log_2m$ ?

Well the minimal sum is increasing with the number of operations, and then it is just a matter of summation by looking at the powers of two that bound the number of operations. We essentially want the sum $\leq mn$. That said, I think it is undeniable that the other solution is much cleaner and nicer.

socrates wrote: Let $n$ numbers all equal to $1$ be written on a blackboard. A move consists of replacing two numbers on the board with two copies of their sum. It happens that after $h$ moves all $n$ numbers on the blackboard are equal to $m.$ Prove that $h \leq \frac{1}{2}n \log_2 m.$ Solved with the help of: @complex_afi $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let's note what happens with the multiplication of all the terms in each operation: $$\ldots\text{ } a \ldots b \text{ } \ldots$$$$\ldots \downarrow \ldots \downarrow \ldots$$$$\ldots (a+b)\ldots (a+b) \ldots$$$$(a+b)^2\ge a^2+b^2+2ab$$By $AM-GM:$ $$(a+b)^2\ge 4ab$$After the $h$ operations: $$\Rightarrow 4^h\le m^n$$$$\Rightarrow log_4 4^h\le log_4 m^n$$$$\Rightarrow h\ge nlog_4 m$$$$\Rightarrow h\ge \frac{1}{2}nlog_2 m_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

Solution: Consider the product of all numbers, each move it at least quadruples, and finally the product is $m^n$. So $m^n \ge 4^h$, giving the result.