<s>[INST] Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$ Let x and a be real numbers > 0. Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$
My idea is that I'm going to use $a>b \iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a} \iff (\frac{1}{2}(x+\frac{a}{x}))^2 = \frac{1}{4}(x+\frac{a}{x})(x+\frac{a}{x})= \frac{1}{4}(x^2+2a+\frac{a^2}{x^2}) \ge \sqrt{a}^2=a$
And from that we get:
$\frac{1}{4}x^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$\frac{1}{4}x^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge \frac{1}{4}a^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge a \iff \frac{1}{4}a^2x^2 +\frac{1}{2}ax^2+\frac{a^2}{4} \ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
 [/INST] From here:
$$\frac{1}{4}x^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge a,$$
move the $a$ to the left side to yield
$$\frac{1}{4}x^2 - \frac{1}{2}a+\frac{a^2}{4x^2} \ge 0.$$
The left side is a perfect square:
$$\left(\frac{x}{2} - \frac{a}{2x}\right)^2.$$
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