Notice first that: \frac{a^3}{(2a^2+b^2)(2a^2+c^2)}=\frac{1}{a}\frac{1}{(2+\frac{b^2}{a^2})(2+\frac{c^2}{a^2})} Since,(2+\frac{b^2}{a^2})(2+\frac{c^2}{a^2})=(1+1+\frac{b^2}{a^2})(1+\frac{c^2}{a^2}+1)\ge(1+\frac{c}{a}+\frac{b}{a})^2=\frac{(a+b+c)^2}{a^2} ...

Step 6 is indeed redundant. One can see that the argument used in Step 6 is essentially identical to that used for proving the 1-1 property of A. Evans might just think it is clearer for students ...

First, \cap_{N=1}^{\infty} ( \cup_{n=N}^{\infty} A_n) = \infty makes no sense as a set can be of infinite cardinality but not be equal to infinity. Now regarding your task: I assume that there is ...

I think the term "median" would be appropriate, though I do not know if anyone else uses this term in this context. Your concept is a generalization of the median of a random variable. This is because ...

Let me try to explain the hint: Suppose you found a y such that Ay \ge 0 and by < 0 and suppose on the contrary that you can find x such that x^TA=b and x \ge 0 . then we have x^TAy=by ...

(see figure 1 below) Let us rewrite the 3 line equations under the equivalent form : (L_1) \ \left\{ \begin{array}[ccc] .(2-z)y &=& zx \\ \ \ z& =& 0 \end{array} \right. , \ \ \ \ \ (L_2) \ \left\{ \begin{array}[rcc] .(2-z)y &= &zx \\ \ \ z& =& 1 \end{array} \right., \ \ \ \ \ (L_3) \ \left\{ \begin{array}[rcc] .(2-z)y& =& zx \\ \ \ z &=& 2 \end{array} \right. ...