djb86 wrote: Let $n \ge 2$ be a fixed even integer. We consider polynomials of the form \[P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + 1\] with real coefficients, having at least one real roots. Find the least possible value of $a^2_1 + a^2_2 + \cdots + a^2_{n-1}$. $x=0$ $ \Rightarrow $ $ P(0)=1$. Answer: min$ (a^2_1+ \cdots +a^2_{n-1})=4 $ For example, $n=2k$ $ \Rightarrow $ $a_i=0$ for all $i \not= k$ and $a_k=-2$.

mathuz wrote: djb86 wrote: Let $n \ge 2$ be a fixed even integer. We consider polynomials of the form \[P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + 1\] with real coefficients, having at least one real roots. Find the least possible value of $a^2_1 + a^2_2 + \cdots + a^2_{n-1}$. $x=0$ $ \Rightarrow $ $ P(0)=1$. Answer: min$ (a^2_1+ \cdots +a^2_{n-1})=4 $ For example, $n=2k$ $ \Rightarrow $ $a_i=0$ for all $i \not= k$ and $a_k=-2$. Wrong. Counter example : $P(x)=x^4+\frac{4}{\sqrt 5}x+1$ $P(-\frac 34)<0$ and so this polynomial has at least one real root and $a_1^2+a_2^2+a_3^2=\frac{16}5<4$ In fact, using the polynomial $x^n-\frac 2{n-1}(x^{n-1}+x^{n-2}+...+x)+1$ (with real root $1$), we get $\sum a_i^2=\boxed{\frac 4{n-1}}$ it's rather easy to show that this is a "local minimum" (any polynomial with another $\sum a_i^2$ always may be slighly modified to decrease the sum). But this is not enough to prove that this is the global minimum for degree $n$ ...

We will show $\frac4{n-1}$ actually is the minimum. $x^n+1$ has no real root, so $a_i$ cannot be all zero. Suppose $r$ is the real root of the polynomial, we have $\sum a_ir^i=-1-r^n$. Now Cauchy inequality implies $\sum a_i^2\geq\frac{ (\sum a_ir^i)^2}{\sum r^{2i}}=\frac{(r^n+1)^2}{\sum r^{2i}}$. It suffices to show $\frac{n-1}4(r^n+1)^2\geq\sum r^{2i}$ We notice $r^n+1\geq r^{n-2i}+r^{2i},\forall 0<i<\frac n2$. Add them together, $\frac{n-2}4(r^n+1)\geq\sum_{i=1}^{n/2-1}r^{2i}$. Multiply by $r^n+1$, $\frac{n-2}4(r^n+1)^2\geq\sum r^{2i} - r^n$. Add $\frac14(r^n+1)^2\geq r^n$, $\frac{n-1}4(r^n+1)^2\geq\sum r^{2i} $. Q.E.D.

xxp2000 wrote: We will show $\frac4{n-2}$ actually is the minimum. But Patrick Collette has already shown that the minimum is at most $\frac 4{n-1}$, and this is smaller than the value $\frac4{n-2}$ claimed by you $\dots$

manuel153 wrote: xxp2000 wrote: We will show $\frac4{n-2}$ actually is the minimum. But Patrick Collette has already shown that the minimum is at most $\frac 4{n-1}$, and this is smaller than the value $\frac4{n-2}$ claimed by you $\dots$ Thanks. It is a typo.

We claim the answer is $\frac{4}{n-1}$ which can be readily verified as the minimum if $x=\pm1$ is a root. Otherwise assume $r\neq 0,\pm1$ is a root. We get $$(a_{n-1}r^{n-1}+\dots+a_1r)^2\leq(a_1^2+a_2^2+\dots+a_n^2)(r^{2n-2}+r^{2n-4}+\dots+r^2)$$$$\frac{(r^n+1)^2}{r^{2n-2}+r^{2n-4}+\dots+r^2}\leq (a_1^2+a_2^2+\dots+a_n^2)$$It is sufficient to show $$4(r^{2n-2}+r^{2n-4}+\dots+r^2)\leq(n-1)(r^{2n}+2r^{n}+1)$$$$4(r^{2n-2}+r^{2n-4}+\dots+r^2)\leq(n-1)((r^{2n}+r^n)+(r^n+1))$$This follows by several applications of the following inequality. If $0\leq a\leq b\leq c\leq d$ and $x>0$ then, $$x^{a+b}+x^{c+d}\geq x^{a+c}+x^{b+d}$$By factoring we get $$(x^a-x^d)(x^b-x^c)\geq 0$$Which is true by spliting into cases on $x$. The result follows.