Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$

Question

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.

Answer 1

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: $$\left(\frac{x}{2} - \frac{a}{2x}\right)^2.$$

Answer 2

Because by AM-GM $$x+\frac{a}{x}\geq2\sqrt{x\cdot\frac{a}{x}}=2\sqrt{a}.$$

Answer 3

$(\sqrt{x}-\sqrt{a/x})^2\geq0\Rightarrow x+a/x-2\sqrt{a}\geq 0\Rightarrow 1/2(x+a/x)\geq \sqrt{a}.$