Suppose that [tex]x,y> 0[/tex] and that [tex]x\leq y[/tex]. We can subtract [tex]x[/tex] from both sides to obtain [tex]y-x \geq 0[/tex]. Recall that [tex](a+b)(a-b)=a^2-b^2[/tex]. Now we can replace [tex]a[/tex] and [tex]b[/tex] in this identity with [tex]\sqrt{y}[/tex] and [tex]\sqrt{x}[/tex] respectively and we get [tex](\sqrt{y}-\sqrt{x})(\sqrt{y}+\sqrt{x})=y-x[/tex]. From this it follows that [tex](\sqrt{y}-\sqrt{x})(\sqrt{y}+\sqrt{x}) \geq 0 [/tex]. Since [tex]\sqrt{y}+\sqrt{x} > 0[/tex], we can divide by it both sides of the inequality without altering its direction and end up with [tex]\sqrt{y}-\sqrt{x} \geq 0[/tex]. Now we just need to add [tex]\sqrt{x}[/tex] to both sides and conclude that [tex]\sqrt{y} \geq \sqrt{x}[/tex], which finishes our proof.
