Prove the Inequality proves true, three variables Prove that for all positive real numbers $a,b,c$ we have:
$$\frac{1}{4a}+\frac{1}{4b}+\frac{1}{4c}+\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\geq\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$$
So far I have solved for $a=b=c=1$ and $a=1$, $b=2$, $c=3$ to show that the inequality holds true but I am needing the answer written more like a proof, I'm just not sure what theorems to apply actually prove the inequality holds true.

Also we can use $uvw$ here.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $\frac{3v^2}{4w^3}+\frac{\sum\limits_{cyc}(3u+a)(3u+b)}{\prod\limits_{cyc}(3u+a)}\geq\frac{\sum\limits_{cyc}(a+b)(a+c)}{9uv^2-w^3}$ or
$\frac{v^2}{4w^3}+\frac{15u^2+v^2}{54u^3+9uv^2+w^3}\geq\frac{3u^2-v^2}{9uv^2-w^3}$, which is $f(w^3)\geq0$, 
where $f$ is a concave function (it's obvious that the coefficient before $w^6$ is negative).
But a concave function gets a minimal value for an extremal value of $w^3$,
which happens in the following cases.


*

*$w^3\rightarrow0^+$. In this case our inequality is obviously true.

*$b=c=1$, which gives $(a-1)^2\geq0$. Done!
