Using the $AM,GM $ : write $x+a_i= (x+1)+1+1+....+1 $ where $1$ is repeated $a_i-1$ time $$x+a_i=(x+1)+ a_i-1 \geq a_i(x+1)^{\frac{1}{a_i}}$$for all $i=1,2,....,n$ multiplying all these inéqualities , we get : $$ (x+a_1)(x+a_2)....(x+a_n) \geq M(x+1)^k$$so $P(x) \leq 0$ for all $x \geq 0$ équality if and only if $x+1=1$ ie $x=0$ . So $P(x)$ has no positive roots .

Just notice that $(x+1)^{1/a_j}\leq 1+\frac{x}{a_j}$ by Bernoulli's inequality, which gives \[P(x)=\prod^n_{k=1}a_k(x+1)^{1/a_k}-\prod^n_{k=1}(x+a_k)\leq \prod^n_{k=1}a_k\left(1+\frac{x}{a_k}\right)-\prod^n_{k=1}(x+a_k)=0.\]Equality occurs if $x=0$ which is not allowed.

We present two solutions, one by calculus and one by AM-GM. First solution (calculus) We will prove the problem in the following form, which lets us carry out the induction. Claim: Relax the conditions on $M$ and $k$ to \[ \frac{1}{a_1} + \dots + \frac{1}{a_n} \le k \in {\mathbb R} \qquad\text{and}\qquad a_1 a_2 \dots a_n = M. \]Then if $x > 0$ we have $P(x) \le 0$ with equality if and only if $M = 1$. Proof. By induction on $n$, with the base case $n=0$ being vacuous (where, $k \le 0$ and $M=1$). Now assume $n \ge 1$. Since $P(0) = 0$ it suffices to prove $P'(x) < 0$, but in fact since $Mk = \sum_{\text{cyc}} a_2 \dots a_n$ we compute \[ P'(x) = \sum_{\text{cyc}} \left[ a_2 \dots a_n (x+1)^{k-1} - (x+a_2)\dots(x+a_n) \right]. \]Each term of the sum is nonpositive by induction hypothesis, and equal to zero only if $a_2 = \dots = a_n = 1$. So $P'(x) = 0$ if and only if $a_1 = \dots = a_n = 1$, or $M=1$. But conversely if $M=1$ we have $P(x) \equiv 0$ anyways. $\blacksquare$ Remark: This actually shows more strongly that all coefficients of $P$ are non-positive, which certainly implies the problem. Second solution (AM-GM) By AM-GM we have \[ x + a_1 = (x+1) + \underbrace{1+\dots+1}_{a_1-1} \ge a_1 \left( x+1 \right)^{\frac{1}{a_1}} \]with equality iff $a_1 = 1$. Multiply cyclically. Remark: Ankan Bhattacharya notes that the (equivalent) inequality $(x+a)^a \ge a^a (x+1)$ is also checked directly by the binomial theorem, so it is not necessary to appeal to AM-GM.

By applying the GM- AM inequality we can easily obtain $M(x + 1)^k = a_1(x + 1)^{1/a_1} * ... * a_n(x + 1)^{1/a_n} = a_1[(x + 1)*1*...*1]^{1/a_1} * ... * a_n[(x + 1)*1*...*1]^{1/a_n} \leq (x + a_1) * ... * (x + a_n) $ (the inequality is applied to each of the $a_i[(x + 1)*1*...*1]^{1/a_i} $ multiples, and the number of 1s in $a_i[(x + 1)*1*...*1]^{1/a_i}$ is $a_i - 1$, for each $ i = {1, ..., n}$), with the equality holding in two cases: Case 1. All of the multiples in each of the $n$ multiples are equal, i.e. $x + 1 = 1$ implying that $x = 0$, which is contradictory to the fact that $x$ is positive. Case 2. $a_i - 1 = 0$ $\forall i = {1, 2, ..., n}$ (meaning the GM- AM inequality was applied n times to only 1 "multiple"), which implies $a_i = 1$ $\forall i = {1, 2, ..., n}$, i.e. $M = a_1 * ... * a_n = 1$, which is also contradictory to the fact that $M > 1$. From these 2 contradictions we obtain that equality never holds, meaning $M(x + 1)^k \gg (x + a_1) * ... * (x + a_n) $, i. e. $P(x) \gg 0$ . $\blacksquare$

This is also Kosovo TST

Let $x > 0$. For each $i$ with $1 \leq i \leq n$ we have by Bernoulli's inequality: $$\left(1 + \frac{x}{a_i}\right)^{a_i} \geq 1 + x \implies x + a_i \geq a_i\left(1 + x\right)^{\frac{1}{a_i}}$$ By multiplying all of these inequalities we find that $$(x + a_1)(x + a_2)\dots(x + a_n) \geq M(1 + x)^k$$ Implying $P(x) \leq 0$. Moreover, the equality in Bernoulli fails to hold whenever $a_i > 1$, which happens at least once because $M > 1$. Thus in fact $P(x) < 0$ for all positive $x$, implying that $P$ has no positive roots, as desired.

By number of simple cyclic inequalities $f^{(k)}(0)\geq 0$ for all $k>1$, so $f(x)=0+0*x+f^{(2)}(0)x^2/2+f^{(3)}(0)x^3/6+\dots \geq 0 $ for $x>0$. If $f(x_0)=0$ for some $x_0>0$, then $f^{(2)}(0)=0$, hence $$\sum_{i\not =j} \frac{1}{a_ia_j} = k(k-1)$$ and all $a_i$ are equal. Inequality $a^n(x+1)^k>(x+a)^n$ is evident Bernoulli. P.S. The last move also helpes to understand how to solve this problem without derivatives.

Note: this solution is clearly the worst solution in the entire thread, but its the first thing that my extremely tired brain could think of, and it works. We will prove the inequality \[(1+x)^{1/a_1+\cdots+1/a_n}<(1+x/a_1)\cdots(1+x/a_n)\]for all $x>0$, and this clearly solves the problem. We will show something even stronger, namely that the inequality holds for the coefficient of $x^j$, so in particular: Lemma: $\binom{1/a_1+\cdots+1/a_n}{j}<\sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_j}$ for $j\ge 2$ (the $j=0$ and $j=1$ terms are 0). Proof of Lemma: We proceed by induction. Equality holds at $j=1$. Therefore, if we showed \[\binom{1/a_1+\cdots+1/a_n}{j-1}\le\sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_{j-1}}\implies\binom{1/a_1+\cdots+1/a_n}{j}<\sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_j},\]we would be done by induction. We will now show this. Note that \begin{align*} \binom{1/a_1+\cdots+1/a_n}{j}&=\frac{1}{j}\binom{1/a_1+\cdots+1/a_n}{j-1}(1/a_1+\cdots+1/a_n-j+1) \\ &< \frac{1}{j}(1/a_1+\cdots+1/a_n-(j-1))\sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_{j-1}}. \end{align*}But \[(1/a_1+\cdots+1/a_n)\sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_{j-1}}=j\sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_j}+\sum_{\mathrm{sym}}\frac{1}{a_1^2\cdots a_{j-1}},\]and since $a_1\cdots a_n>1$, we see that \[\sum_{\mathrm{sym}}\frac{1}{a_1^2\cdots a_{j-1}}<(j-1)\sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_{j-1}}.\]Thus, \begin{align*} \binom{1/a_1+\cdots+1/a_n}{j} &\le \frac{1}{j}(1/a_1+\cdots+1/a_n-(j-1))\sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_{j-1}} \\ &= \sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_j}+\frac{1}{j}\sum_{\mathrm{sym}}\frac{1}{a_1^2\cdots a_{j-1}}-\frac{j-1}{j}\sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_{j-1}} \\ &< \sum_{\mathrm{sym}}\frac{1}{a_1\cdots a_j}, \end{align*}as desired, so the lemma is proven by induction on $j$. $\blacksquare$ This also completes the proof of the problem.

My solution: We start with the following lemma. Lemma Suppose $m \in \mathbb{N}$ and $x \in \mathbb{R^+} \cup \{0\}$. Then we have $$x+m \geq m(x+1)^{\frac{1}{m}}$$ PROOF: By Bernoulli's inequality, we have $$\left(1+\frac{x}{m} \right)^m \geq 1+x$$Taking $\frac{1}{m}^{th}$ power on both sides and then multiplying by $m$, we get the Lemma. $\Box$ Return to the problem at hand. Taking $m=a_1,a_2, \dots ,a_n$ in our lemma, and multiplying the obtained inequalities, we have $(x+a_1)(x+a_2) \dots (x+a_n) \geq M(x+1)^k$. Thus, as $M >1$, so $P(x)=0 \Rightarrow (x+a_i)^{a_i}=a_i(x+1) \text{ } \forall i \in \{1,2, \dots ,n\}$ $\Rightarrow x=0$ is the only non-negative real solution to $P(x)=0$. Therefore, $P(x)=0$ has no solution in positive reals. $\blacksquare$

It suffices to prove for positive $x$ $$a_i^{a_i}*(x+1) \le (x+a_i)^{a_i}$$with equality iff $a_i = 1$ Now, Binomial Theorem and chill.

This is a good problem,with a short,simple proof. My proof is as follows: Assume the contrary,that it has some positive root $x$. Then,$x+a_i = x+1+1+1+1...... \geq a_i(x+1)^{\frac{1}{a_i}}$,where the 1's are repeated $a_i-1$ times.Note that equality will not occur as $x>0$.Multiply this for all $1 \leq i \leq n$ to get $P(x)=M(x+1)^k-(x+a_1)(x+a_2)\cdots (x+a_n) <0$,which is a contradiction to our assumption.Hence there exist no such root.$\blacksquare$

We will show that $\prod_{i = 1}^{n} (x + a_i) > M(x + 1)^k$ for all $x > 0$. This is sufficient in showing that $P(x)$ has no positive roots. The RHS becomes $\prod_{i = 1}^{n} a_i (x+1)^{\frac{1}{a_i}}$. Now $x + a_i > a_i(x + 1)^{\frac{1}{a_i}}$ gives $(1 + x)^{\frac{1}{a_i}} < 1 + \frac{x}{a_i}$ which is true by Bernoulli, so $\prod_{i = 1}^{n} (x + a_i) > \prod_{i = 1}^{n} a_i(x+1)^{\frac{1}{a_i}} = M(x+1)^k$ as desired. $\square$

Notice that the polynomial having roots is equivalent to having $(x+a_1)...(x+a_n)=(a_1(1+x)^{\frac 1{a_1}}) \cdot ... \cdot (a_n (1+x)^{\frac 1{a^n}})$. But notice $(1+\frac{x}{a_i})^{a_i} = 1+ \binom{a_i}{1} \frac{x}{a_i}+... \geq 1+x=>(x+a_i) \geq a_i (1+x)^{\frac 1{a_1}}$, where equality holds iff $a_i=1$. As all $a_i$ can't be 1 as $M>1$, the statement follows. Cute A1 , and the way of stating at least one $a_i$ was greater than $1$ was kind of a throw-off; also, it need not have been a polynomial for it doesn't affect the inequality at all, but the problem seems to have been built to try it's best to throw you down the wrong path when you first see it

Suppose that $P(r)=0$ for some $r>0$. Observe that by AM-GM, we have $$r+a_i = (r+1)+\underbrace{1+1+\cdots+1}_{a_i-1} \ge a_i(r+1)^{\tfrac{1}{a_i}}$$for any positive integer $a_i$. Then \begin{align*}P(r) &= M(r+1)^k - (r+a_1)(r+a_2) \cdots (r+a_n) \\ &= \prod_{i=1}^{n} a_i(r+1)^{\tfrac{1}{a_i}} - \prod_{i=1}^{n} (r+a_i) \\ &= \prod_{i=1}^{n} [a_i(r+1)^{\tfrac{1}{a_i}}-(r+a_i)] \\ &\le 0.\end{align*}Equality occurs only if $r+1=1$ or $a_1 = a_2 = \cdots = a_n = 1$; but $r>0$ and $M>1$, contradiction. We can conclude that $P(x)$ has no positive real root, as desired. $\blacksquare$

We will show that (x+a1)(x+a2)...(x+an) > M(x+1)^k for all positive real numbers x; First notice that M(x+1)^(k)=a1(x+1)^(1/a1) * a2(x+1)^(1/a2) *...* an(x+1)^(1/an). Now using AM-GM we have ai(x+1)^(1/ai)=< (x+1)+1+1+...+1=x+ai. ai-1 times Now combining all these together we get (x+a1)(x+a2)...(x+an)>M(x+1)^(k) wich means that P(x) is negative for all positive real numbers x i.e. P(x) has no positive real roots.

As we all know by the $AM-GM$ inequality $$ x+a_i = (x+1)+1+1+....+1 \geqslant a_i(x+1)^{\frac{1}{a_i}} $$Where $x$ must be a positive real,and where $i$,is any $\mathbb{N}$ in the interval $i \in [1,n] $ The equality case holds only when $x+1=1 \implies x=0$,but $x>0$. Thus for every positive real $x$ we have that the polynomial $P(X)$ is: $$ P(x) = M(x+1)^k-(x+a_1)(x+a_2)...(x+a_n) < M(x+1)^k-a_1(x+1)^{\frac{1}{a_1}}...a_n(x+1)^{\frac{1}{a_n}} = M(x+1)^k-a_1a_2...a_n(x+1)^{\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}}=M(x+1)^k-M(x+1)^k=0 $$Therefore for every positive real number x the polynomial $P(x) < 0$ Which means that it doesn't have any positive real roots

solution which requires no cleverness: let $f(x)=\sum_{i=1}^{n} \log (x+a_i)-\log (M) -k\log (x+1)$. Now $f(0)=0$. If $f(a)=0$ for $a>0$, then by rolle there exists $a'>0$ such that $f'(a')=0$. Which implies $$\sum_{i=1}^{n} \dfrac{1}{a'+a_i}=\left (\sum_{i=1}^n \dfrac{1}{a_i} \right )\dfrac{1}{a'+1}$$. But this is not possible bcoz $\dfrac{1}{a'+a_i} \ge \dfrac{1}{a_i(a'+1)}$ and since $M>1$, the ineq is strinct for some $i$. Summing these we find LHS>RHS, contradiction.

Lemma: For any positive integer $a>1$ we have $x+a > a(x+1)^{\frac{1}{a}}$ for all $x>0$. Proof: Clear by raising both sides to the power of $a$ and using binomial expansion. Thus it follows that $(x+a_1) \cdots (x+a_n) > M(x+1)^k$ as desired.

We can rewrite $P(x)$ as: \[\prod_{i=1}^{n} a_i(x+1)^{\frac{1}{a_i}}-\prod_{i=1}^{n}(x+a_i).\]Claim: $\forall$ $x>0$, $P(x)<0$, which implies the desired result Proof: By AM-GM: \[x+a_i\geq a_i(x+1)^{\frac{1}{a_i}},\]on the domain of $x>0$. However, equality occurs iff $x=0$, implying that: \[x+a_i>a_i(x+1)^{\frac{1}{a_i}}.\]This implies that $P(x)<0$, as desired $\square$

We claim that for any $x>0$, we have that \[\left(x+a_1\right)\cdot\left(x+a_2\right)\cdots\left(x+a_n\right)>M\left(1+x\right)^k\]\[\Longleftrightarrow \left(1+\frac x{a_1}\right)\cdot\left(1+\frac x{a_2}\right)\cdots\left(1+\frac x{a_n}\right)>\left(1+x\right)^k\]Note that for every $1\leq i\leq n$, applying Bernoulli's Inequality, we have \[\left(1+\frac x{a_i}\right)^{a_i}>1+x\Longrightarrow \left(1+\frac x{a_i}\right)>\left(1+x\right)^{\frac 1{a_i}}\]Multiplying all the $n$ inequations, we get \[\left(1+\frac x{a_1}\right)\cdot\left(1+\frac x{a_2}\right)\cdots\left(1+\frac x{a_n}\right)>\left(1+x\right)^{\frac 1{a_1}+\frac 1{a_2}+\cdots+\frac 1{a_n}}=\left(1+x\right)^k\]Hence, for every $x>0$, we have $P(x)<0$. So, $P(x)=0$ is impossible for $x>0$, as desired. $\blacksquare$

Assume $x > 0.$ By Bernoulli's Inequality, for all $1 \le i \le n,$ we have \[ \left(1 + \frac{x}{a_i}\right)^{a_i} \ge 1 + x. \]Taking the $a$th root of both sides preserves the inequality, giving \[ 1 + \frac{x}{a_i} \ge (1+x)^{\frac{1}{a_i}}.\]Mutiplying both sides by $a_i$ (since $a_i > 0$) also preserves the inequality, giving \[ \boxed{x + a_i \ge a_i (1+x)^{\frac{1}{a_i}}}. \]Since $x > 0,$ both sides of this expression are positive. Thus multiplying the boxed expression across all $i$ gives \[ M(x+1)^k \le (x+a_1)(x+a_2) \dots (x+a_n). \]Therefore, $P(x) \le 0$ for all $x > 0.$ Equality holds if and only if the equality holds in all instances of Bernoulli's inequality. This happens if and only if either $x = 0$ or $a_i = 1$ for all $i.$ The former can't hold by our initial assumption, and neither can the latter since that would impose $M = 1,$ contradicting the problem statement. Thus equality never holds, and so $P(x) < 0$ for all $x > 0.$ Therefore, $P(x)$ has no positive roots, as desired.

It suffices to show that $\forall x > 0, M(x+1)^k - \prod_{i=1}^{n}(x+a_i) < 0$. Lemma: ${(x+1)}^{\frac{1}{p}} < 1+\frac{x}{p}$ for any positive integer $p > 1$ and positive number $x$. Moreover, when $p = 1$, ${(x+1)}^{\frac{1}{p}} = 1+\frac{x}{p}$ Note that $(1+\frac{x}{p})^p = 1 + \binom{p}{1}\frac{x}{p} + \binom{p}{2}(\frac{x}{p})^2 + \binom{p}{3}(\frac{x}{p})^3 \dots > 1 + x$. So it follows that $(1+\frac{x}{p}) > {(1+x)}^{\frac{1}{p}}$ by taking the $p$th root of both sides. Since $M > 1$, we know that $n > 1$ and there is some $j$ such that $a_j > 1$. So $\prod_{i=1}^{n}{(1+x)}^{\frac{1}{a_i}} < \prod_{i=1}^{n} (1+\frac{x}{a_i}) \implies (1+x)^{k} < \prod_{i=1}^{n} (1+\frac{x}{a_i}) = \prod_{i=1}^{n} (\frac{x+a_i}{a_i}) \implies {(1+x)}^{k}\prod_{i=1}^{n}a_i < \prod_{i=1}^{n}(x+a_i) \implies M{(1+x)}^{k} < \prod_{i=1}^{n}(x+a_i)$ when $x > 0$

FTSOC we assume that $\exists$ $\alpha \in \mathbb{R}^{+}$ such that $P(\alpha)=0$ , we notice $P(0)=0$ so , $\exists$ $\xi \in (0, \alpha)$ s.t. $P'(\xi)=0$ , Now $P'(x)=\sum_{\mathrm{cyc}} \left(a_1a_2\cdots a_r(x+1)^{k-1}-(x+a_1)(x+a_2)\cdots(x+a_r)\right)$ , since $\frac{1}{a_i} \leqslant 1 \implies k \leqslant n$ , and hence we have $P'(x) \leqslant 0$ , it is $0$ when $a_i=1$ $\forall$ $i$ , but that will force $M=1$ , which is not true hence $P'(x)<0$ , but that contradicts , $P'(\xi)=0$ , Hence $\nexists$ $\alpha \in \mathbb{R}^{+}$ s.t. $P(\alpha)=0$. $\square$ Another solution (clever) is: for $x>0$ we have $\frac{(x+1)+\underbrace{(1+1+1+\cdots+1)}_{\mathrm{a_{i}-1\hspace{0.1cm} terms}}}{a_{i}} \geqslant \sqrt[a_{i}]{(x+1)} \implies x+a_{i} \geqslant a_{i}(x+1)^{\frac{1}{a_{i}}}$. Now we notice $$\prod_{i=1}^{i=n} a_{i}(x+1)^{\frac{1}{a_{i}}}=M(x+1)^{k} \leqslant \prod_{i=1}^{i=n}(x+a_{i}) \implies P(x) \leqslant 0$$, with equality iff $a_{i}=1$ which is not true since $M>1$ , hence $P(x)<0$ for all $x \in \mathbb{R}^{+}$. $\square$

v_Enhance wrote: We present two solutions, one by calculus and one by AM-GM. First solution (calculus) We will prove the problem in the following form, which lets us carry out the induction. Claim: Relax the conditions on $M$ and $k$ to \[ \frac{1}{a_1} + \dots + \frac{1}{a_n} \le k \in {\mathbb R} \qquad\text{and}\qquad a_1 a_2 \dots a_n = M. \]Then if $x > 0$ we have $P(x) \le 0$ with equality if and only if $M = 1$. Proof. By induction on $n$, with the base case $n=0$ being vacuous (where, $k \le 0$ and $M=1$). Now assume $n \ge 1$. Since $P(0) = 0$ it suffices to prove $P'(x) < 0$, but in fact since $Mk = \sum_{\text{cyc}} a_2 \dots a_n$ we compute \[ P'(x) = \sum_{\text{cyc}} \left[ a_2 \dots a_n (x+1)^{k-1} - (x+a_2)\dots(x+a_n) \right]. \]Each term of the sum is nonpositive by induction hypothesis, and equal to zero only if $a_2 = \dots = a_n = 1$. So $P'(x) = 0$ if and only if $a_1 = \dots = a_n = 1$, or $M=1$. But conversely if $M=1$ we have $P(x) \equiv 0$ anyways. $\blacksquare$ Remark: This actually shows more strongly that all coefficients of $P$ are non-positive, which certainly implies the problem. Second solution (AM-GM) By AM-GM we have \[ x + a_1 = (x+1) + \underbrace{1+\dots+1}_{a_1-1} \ge a_1 \left( x+1 \right)^{\frac{1}{a_1}} \]with equality iff $a_1 = 1$. Multiply cyclically. Remark: Ankan Bhattacharya notes that the (equivalent) inequality $(x+a)^a \ge a^a (x+1)$ is also checked directly by the binomial theorem, so it is not necessary to appeal to AM-GM. That's an amazing induction

Working backwards, we have \begin{align*} M(x+1)^k = \prod \left( a_i (x+1)^{\left( \tfrac{1}{a_i} \right)} \right) &< \prod (x+a_i) \\ \prod (x+1)^{\left( \tfrac{1}{a_i} \right)} &< \prod \left( \frac{x}{a_i} + 1 \right). \end{align*}In fact, each term on the LHS is less than its corresponding term on the RHS: \begin{align*} (x+1)^{\left( \tfrac{1}{a_i} \right)} &\leq \left( \frac{x}{a_i} + 1 \right) \\ x+1 &\leq \left( \frac{x}{a_i} + 1 \right)^{a_i} \end{align*}which is true by Bernoulli's inequality; the equality case will not hold for whichever $a_i$ is not equal to $1$.

Claim: $a_i(x+1)^{\frac{1}{a_i}} \le (x+a_i) \forall x > 0$. Proof: Observe that $$a_i(x+1)^{\frac{1}{a_i}} \le (x+a_i) \iff (x+1)^{\frac{1}{a_i}} \le (1 + \frac{x}{a_i}) \iff (1 + \frac{x}{a_i})^{a_i} \ge x+1,$$which is true due to the Binomial Theorem. Note that in particular, equality holds iff $a_i = 1$. $\square$ Note that due to Claim, $$\prod_i a_i(x+1)^{\frac{1}{a_i}} \le \prod_i (x+a_i).$$But for equality to hold here, equality has to hold everywhere, i. e. $a_i = 1 \forall i$. But then $1 > M = 1$, contradiction. Therefore, $\prod_i a_i(x+1)^{\frac{1}{a_i}} < \prod_i (x+a_i) \iff M(x+1)^k<(x+a_1)(x+a_2)\cdots (x+a_n) \iff P(x) = M(x+1)^k-(x+a_1)(x+a_2)\cdots (x+a_n) < 0 \forall x$ Since $P(x)$ is always negative for positive $x$, it can have no positive root. $\square$

We need to prove that \begin{align*}P(x)&=(a_1a_2\cdots a_n)(x+1)^{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}\\&\leq (x+a_1)(x+a_2)\cdots(x+a_n)\end{align*}By AM-GM we have $$x+a_i=(x+1)+ 1(a_i-1) \geq a_i(x+1)^{\frac{1}{a_i}}$$multiplying for all $a_i$ we get the above equation.

By AM-GM, \[\frac{x+a_i}{a_i}=\frac{(x+1)+1+\dots +1}{a_i}\geq (x+1)^{\frac{1}{a_i}}.\]Equality holds iff $x=0$ or $a_i=1$, so equality must not hold for some $i$ as $M>1$. Multiplying these finishes

Consider the polynomial $\prod (x + a_i) - M(x + 1)^k$. Since $n > k$, this goes to infinity. Assume then that there exists some positive root, then at the root with the largest value, the derivative of the function must be nonnegative at this root. The derivative $y = \prod (x + a_i)$ can be given by taking logarithms and then writing $\frac{\frac{\mathrm{d} y}{\mathrm{d} x}}{y} = \sum \frac{1}{x + a_i}$, so the desired derivative is $(\sum \frac{1}{x + a_i})\prod (x + a_i) $, and the derivative of the whole function is $\sum{ \frac{1}{x + a_i}} \prod(x + a_i) - k(x + 1)^{ k - 1}$. We then require $ \sum{ \frac{1}{x + a_i}} \prod(x + a_i) \ge Mk (x + 1)^{k - 1}$, dividing by $(x + 1)^k = \prod (x + a_i)$ forces $\sum \frac{x + 1}{x + a_i} \ge Mk$, but $Mk = \sum_{1 \le i \le n} \prod_{i \neq j} a_j \ge \sum_{1 \le i \le n} 1 = n$, but this is clearly impossible since $\sum \frac{x + 1}{x + a_i} \le n$, so we are done.

lol, the hints are just too obvious

$$x+a_1 = (x+1)+\underbrace{1+\dots+1}_{a_1-1} \geq a_1 (x+1)^\frac{1}{a_1}$$by AMGM, multiplying cyclically finishes. Equality holds if $x=0$ (contradiction) or $a_1=1$. But $M > 1$, so we are done.

By binomial expansion we have for all $x>0$ and integers $1\le i\le n$ that $x+1\le(\frac{x}{a_i}+1)^{a_i}$ and thus equivalently that \[a_i(x+1)^{\frac1{a_i}}\le x+a_i,\]with equality if and only if $a_i=1$. Multiplying the above over all values of $i$ then yields that $M(x+1)^k\le(x+a_1)(x+a_2)\dots(x+a_n)$, but equality is impossible at it forces $a_i=1$ for all $i$, giving $M=1$. Hence the inequality is strict, so that $P(x)<0$ for all $x>0$. $\blacksquare$

Suppose for the sake of a contradiction that $x$ is a positive root. By Bernoulli's Inequality, $$\left( \frac{x}{a_i} + 1 \right) \geq (1+x)^{1/a_i}.$$Therefore $$P(x) = M(1+x)^k - (x+a_1)(x+a_2)(\cdots)(x+a_n) \leq M(1+x)^k-M(1+x)^{1/a_1+1/a_2+\cdots+1/a_n} = 0.$$But equality holds if either $x=0$ or all $1/a_i=1.$ The former is clearly not true, while the latter is also impossible because $M > 1.$ Therefore the inequality is strict, a contradiction. QED