Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$ Question:

Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$.

So for this question, I began by expanding all terms and moving them all to one side. However, I do not know how to definitively say that the statement is proved.
This is my "work" so far:
$a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$
$\frac {2a(b + 2a) + 2b(a + 2b) - 1(b + 2a)(a + 2b)}{2(a + 2b)(b+2a)} ≥ 0$
$\frac {4ab + 4a^2 + 4b^2 - 2b^2 - 5ab - 2a^2}{4b^2 + 10ab + 4a^2} ≥ 0$
$\frac {2a^2 - ab + 2b^2}{4b^2 + 10ab + 4a^2} ≥ 0$
 A: You almost have it, just use
$$2a^2-ab+2b^2 = 2(a-b)^2+3ab > 0$$
(and the denominator is positive as well).
A: One of $a$ or $b$ is non zero, or else the question is not well defined.
If $a=b$, then we get left hand side of inequality to be $2/3$, which is $\geq 1/2$ as desired.
So assume $a>b$. Then
$$ \frac{a}{a+2b} + \frac{b}{b + 2a} 
\\ = \frac{a(b+2a)+b(a+2b)}{(a+2b)(b+2a)} 
\\ = \frac{2ab+2a^2+2b^2}{3ab + (2ab+2a^2 + 2b^2)}
\\ = \frac{x}{3ab + x} \mbox{ where } x = 2ab+2a^2 + 2b^2 > 0
\\ = \frac{1}{\frac{3ab}{x} + 1}
$$
So we need to show that
$$\frac{3ab}{x} \leq 1 \iff 3ab \leq x \iff ab \leq 2a^2+2b^2 $$
But the furthest right inequality is obvious  since
$$ ab < a^2 < 2a^2 < 2a^2 + 2b^2 $$
A: first let's do some cleaning:
$$\frac{a}{a+2b}+\frac{b}{b+2a} = \frac{2a^2 + 2ab + 2b^2}{2a^2 + 4ab + 2a^2} = \frac{2(a+b)^2 - 2ab}{2(a+b)^2} = 1-\frac{ab}{(a+b)^2}$$
and we have :
$$A = \frac{a}{a+2b}+\frac{b}{b+2a} = \frac{1}{1+\frac{2b}{a}} + \frac{1}{1+\frac{2a}{b}}$$
let $x = \frac{a}{b} $ then we have :
$$f(x) = \frac{1}{1+\frac{2}{x}} + \frac{1}{1+2x} = \frac{x}{x+2} + \frac{1}{1+2x}$$
therefore:
$$f^\prime(x) = \frac{2}{(x+2)^2} - \frac{2}{(1+2x)^2} = 0 \Rightarrow x^2 + 4x + 4 = 4x^2 + 4x + 1\Rightarrow 3x^2-3 = 0 \Rightarrow x = \pm 1 \Rightarrow f(x) = \frac{2}{3}, -2$$
but since both a and b are positive x = -2 is not an option and therefore we have that with x>0 f(x) is always larger than 2/3 which is a subset of the desired result.$\blacksquare$
