7) Which of the inequalities below are satisfied for every positive real value of $x$?
a) $x^{2}+\frac{1}{x^{2}} \geq 2$

(A) All three (B) only
b) $x^{2} \geq 2 x-2$

(D) only a) (E) only c).

(C) only b) and c)
c) $x^{4}-2 x^{3}+x^{2} \geq 0$.
7) The answer is (A). a) is equivalent to $x^{4}-2 x^{2}+1=\left(x^{2}-1\right)^{2} \geq 0$, b) is equivalent to $x^{2}-2 x+2=$ $(x-1)^{2}+1 \geq 0$, c) is $\left(x^{2}-x\right)^{2} \geq 0$.

SECOND SOLUTION.

All three given inequalities are satisfied for every positive value of $x$. Indeed,

$x^{2}+\frac{1}{x^{2}}-2=\left(x-\frac{1}{x}\right)^{2}$ and a square cannot be negative.

$x^{2}-2 x+2=(x-1)^{2}+1$ is a sum of squares and therefore is always positive.

$x^{4}-2 x^{3}+x^{2}=x^{2}(x-1)^{2}$ is the product of two squares and therefore cannot be negative.