Let \[ P(x) = \sum_{i=1}^{N_1}a_i x^i\quad\text{and}\quad Q(x) = \sum_{j=1}^{N_2} b_jx^j, \]using $P(0)=Q(0)=0$. (1) Since $Q(x)$ has non-negative coefficients, $Q(x)\ge 0$ on $[0,1]$ is clear. Now, for any $x\in[0,1]$ and $j\ge 1$, $x^j\le x$, forcing $Q(x)\le \textstyle \sum_{1\le j\le N_2}b_j x = xQ(1)\le x$, as claimed. Lastly, $P'(0) = a_1\ge 1$ yields together with $a_i\ge 0$ that $P(x)\ge a_1 x\ge x$ for any $x\in[0,1]$. This completes part 1. (2) The inequality holds for $x=0$, so let $x\in(0,1]$. It's not hard to see that \[ Q(P(x)) = \sum_{1\le j\le N_2}b_j P(x)^j = P(x)\sum_{1\le j\le N_2} b_j P(x)^{j-1}\ge P(x)\sum_{1\le j\le N_2} b_j x^{j-1} = \frac{P(x)Q(x)}{2}, \]using $P(x)\ge x$ per part 1. Likewise, $P(Q(x))\le P(x)Q(x)/2$, using $Q(x)\le x$, yielding the conclusion.

Iveela wrote: Let $P(x)$ and $Q(x)$ be polynomials with nonnegative coefficients. We denote by $P'(x)$ the derivative of $P(x)$. Suppose that $P(0)=Q(0)=0$ and $Q(1) \leq 1 \leq P'(0)$. $(1)$ Prove that $0 \leq Q(x) \leq x \leq P(x)$ for all $0 \leq x \leq 1$. $(2)$ Prove that $P(Q(x)) \leq Q(P(x))$ for all $0 \leq x \leq 1$. Proposed by Otgonbayar Uuye. it's joever

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