Let $ p(x)=x(x-1)\dots(x-20)$. Then we have to prove that if $ p(a_1)+p(a_2)+\dots+p(a_20) \leq 100.99.\dots.79$ then $ a_1+a_2+\dots a_20 \leq 9900$. Now $ p(x)=0 \forall x=0,1,\dots,20$, and if $ a \geq b \geq 20$ then since $ a-i \geq b-i \geq 0 \forall 0 \leq i \leq 20$, $ p(a) \geq p(b) \geq 0 = p(20)=p(19)=\dots=p(0)$, and $ a(a-1)\dots(a-19) \geq b(b-1)\dots(b-19)$. Therefore $ p(x)$ is increasing on the nonnegative integers. Also, if $ a>b$ are nonnegative integers, then $ p(a)+p(b)\geq p(a-1)+p(b+1)$ $ \iff (a-1)(a-2)\dots(a-20)[a-(a-21)] \geq b(b-1)\dots[b+1-(b-20)] %Error. "doubleleftarrow" is a bad command. a-1 \geq b$ is true since $ a>b$. Hence we can always take the most and least of the $ a_i$ and move them closer together while keeping the sum of the $ a_i$ constant. We continue doing this until the $ a_i$ consist of exactly two consecutive nonnegative integers: $ a$ with multiplicity $ x$ and $ a+1$ with multiplicity $ 20-x$. Then we have $ xp(a) +(20-x)p(a+1) \leq 100.99.\dots.79$.

Suppose that there exist $ a_i \geq 100$ , then $ a_i(a_i - 1)(a_i - 2) ...(a_i - 20) > 100 \cdot 99 ...\cdot 79$ ,contradiction. Thus all $ a_i$ are less or equal than $ 99$ . This imply $ a_1 + a_2 + ... + a_{100} \leq 99 \cdot 100 = 9900$ ,as desired

enndb0x wrote: Suppose that there exist $ a_i \geq 100$ , then $ a_i(a_i - 1)(a_i - 2) ...(a_i - 20) > 100 \cdot 99 ...\cdot 79$ ,contradiction. Thus all $ a_i$ are less or equal than $ 99$ . This imply $ a_1 + a_2 + ... + a_{100} \leq 99 \cdot 100 = 9900$ ,as desired Definitely not.

not so hard and dantern. it can be second-killed by noting that polynomials $C_x^n$ are convex functions.

littletush wrote: not so hard and dantern. it can be second-killed by noting that polynomials $C_x^n$ are convex functions. How is taht solution with $C_x^n$ convex functions

If we know that Cnx is convex then f(x) = x!/(x-21)! Is convex meaning we know f(a1)+f(a2) + ... + f(a100) <= 100*f(99) and then it is easy to finish with Jensen's inequality.

Sorry to revive but how could it be proved that $\binom{x}{n}$ is convex?

XX-math-XX wrote: Sorry to revive but how could it be proved that $\binom{x}{n}$ is convex? Well, what we need is that it is convex for $x \ge n$. But its second derivative is a polynomial of degree $n-2$ and by Rolle, it has $n-2$ roots in $[0,n]$ (since $\binom{x}{n}$ has all its $n$ roots there), hence must be of constant sign and hence positive for $x\ge n$.