I posted the question before show the quadratic function is quasi-concave. But I can not understand the only answer, and the author delated his account already. I rephrase my question and show my second attempt here. I have a quadratic function $W(x_1,x_2,\ldots,x_n)=A\sum_{i} x_i^2+ \sum_{i\neq j} x_ix_j$, with $x_i$ nonnegative and $A \in[0,1)$. And w.l.o.g. we can normalize $x_i's$ to between 0 and 1. In quadratic form, it is $W(x_1,x_2,\ldots,x_n)=X^TMX$, where $$M=\begin{bmatrix} A & 1 & 1 & \dots & 1 \\ 1 & A & 1 & \dots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \dots & A \end{bmatrix}.$$ I want to show it is quasi-concave. First I checked $W$ is not concave or convex as $M$ is indefinite. A sufficient condition for (strictly) quasi-concavity of $W$ is that $$(-1)^kH_k(x)>0$$ for $k=1,2...,n.$ Where $H_k(x)$ is the $k-th$ order leading principal minor of the bordered Hessian matrix of $W(x)$:$$H(x)=\begin{bmatrix} 0 & 2A+2\sum_{i\neq1}x_i & 2A+2\sum_{i\neq2}x_i & \dots & 2A+2\sum_{i\neq n}x_i \\ 2A+2\sum_{i\neq1}x_i & 2A & 2 & \dots & 2 \\ 2A+2\sum_{i\neq 2}x_i & 2 & 2A & \dots & 2\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2A+2\sum_{i\neq n}x_i & 2 & 2 & \dots & 2A \end{bmatrix}.$$ The determinant of this juggernaut (and all its leading principle minors) can be decomposed. $$H_k(x)=\begin{bmatrix} 0 & 2A+2\sum_{i\neq1}x_i & 2A+2\sum_{i\neq2}x_i & \dots & 2A+2\sum_{i\neq k}x_i \\ 2A+2\sum_{i\neq1}x_i & 2A & 2 & \dots & 2 \\ 2A+2\sum_{i\neq 2}x_i & 2 & 2A & \dots & 2\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2A+2\sum_{i\neq k}x_i & 2 & 2 & \dots & 2A \end{bmatrix}.$$ Let the first row except leading 0 to be $D$, the first column except leading 0 to be $C$, denote the inner $k*k$ sub-matrix to be$$B=\begin{bmatrix} 2A & 2 & 2 & \dots & 2 \\ 2 & 2A & 2 & \dots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 2 & 2 & 2 & \dots & 2A \end{bmatrix}.$$ Which turns out to be 2 times of matrix $M$. Then $$det(H_k(x))=det(B)det(0-DB^{-1}C).$$ I know the pattern of $det(B).$ to know whether the sufficient condition is satisfied, I just need to know the sign of $-DB^{-1}C$ which is a number. But as $B$ is indefinite, $B^{-1}$ is indefinite also, and I actually tried to write out $-DB^{-1}C$ but it is long, and I can not see a easy answer. Is the fact $2A+2\sum_{i\neq j}x_i<2n$ for any $j$ helpful?
Can anyone explain the answer in the original post? Or help me to find a way out here? Checking the bordered Hessian is a standard textbook method, but economist normally find shortcut specific to the context, however I am not able to find one.
This is my 3rd attempt:
But it seems we can not say $XY >=Y^2$ because we do not know the shape of $W(X)=W(Y)=C$, we are yet to prove it. It is possible the projection of $X$ onto $Y$ is shorter than $Y$ even though we assume $X$ is longer than $Y$.
We have with $\alpha=1-A\in (0,1]$: $$ M = -\alpha \ 1_n + e e^T$$ where $e^T=[1\ 1 \ ... \ 1]$. We set $$B(x,y)= x^T M y= -\alpha x^T y + (e^Tx)(e^Ty)\ \ \ \mbox{and} \ \ \ W(x)=B(x,x).$$
We wish to show that $W(x)$ is strictly quasi-concavity on $\Omega={\Bbb R}_+^n$. We note that $W$ is positive on $\Omega$. It suffices to show that if the conditions $x\neq y\in \Omega$ and $1=W(x)=W(y)$ hold then for all $t\in (0,1)$: $W(tx+(1-t)y)- 1=2t(1-t)(B(x,y)-1)>0$. So assume these conditions. We have (for $x_i,y_i\geq 0$): $$ 2x^T y < x^Tx + y^T y \ \ \mbox{and} \ \ (x^T y)^2 \leq (x^Tx) \ (y^T y)$$ which implies the inequality: $$ (e^T x)^2 (e^T y)^2 =(1 + \alpha x^T x) (1+ \alpha y^Ty) > (1 + \alpha x^T y)^2 $$ And from this we have: $$ (e^Tx)(e^Ty) > 1+ \alpha x^Ty \ \ \mbox{or} \ \ B(x,y)>1 $$ which is what we wanted to show.