First of all, let's prove that
[tex]x+\dfrac{1}{x}\geq 2,\quad x>0[/tex]
We can rewrite the fraction as
[tex]\dfrac{x^2+1}{x}[/tex]
And the inequality as
[tex]\dfrac{x^2+1}{x}-2\geq 0[/tex]
Again, we can rewrite as
[tex]\dfrac{x^2-2x+1}{x}\geq 0 \iff \dfrac{(x-1)^2}{x}\geq 0[/tex]
The numerator is a square, so it is always greater than or equal to zero.
The denominator is positive, because we're assuming x>0.
So, this is a fraction between positive numbers, and it is always positive, and the first part is finished.
As for the second part, we have
[tex]x^2+\dfrac{1}{x^2} \geq x+\dfrac{1}{x} \iff x^2+\dfrac{1}{x^2}-x-\dfrac{1}{x} \geq 0 \iff \dfrac{(x - 1)^2 (x^2 + x + 1)}{x^2} \geq 0[/tex]
The numerator is not the product of a square (always greater than or equal to zero) and a quadratic equation with no solutions and positive leading coefficient (always positive), so it's always greater than or equal to zero. The denominator is x^2, and since x>0 by hypothesis, it is positive. This ends the proof.
