Prove the Inequality proves true, three variables

Question

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.

Answer 1

For all real numbers it's wrong, of course.

For positives $a$, $b$ and $c$, we can use an Integral method:

We see that for all positives $x$, $y$ and $z$ the following inequality holds. $$\sum\limits_{cyc}(x^4-2x^2y^2+x^2yz)\geq0$$ Indeed, by Schur $$\sum\limits_{cyc}(x^4-2x^2y^2+x^2yz)=\sum\limits_{cyc}(x^4-x^3y-x^3z+x^2yz)+\sum\limits_{cyc}xy(x-y)^2\geq0$$ Hence, $\sum\limits_{cyc}\left(t^{4a}-2t^{2a+2b}+t^{2a+b+c}\right)\geq0$, where $t>0$ or

$\sum\limits_{cyc}\left(t^{4a-1}-2t^{2a+2b-1}+t^{2a+b+c-1}\right)\geq0$.

Thus, $\int\limits_{0}^1\sum\limits_{cyc}\left(t^{4a-1}-2t^{2a+2b-1}+t^{2a+b+c-1}\right)dt\geq0$,

which gives your inequality. Done!

Answer 2

Also we can use a full expanding: $$\sum\limits_{sym}(2a^6b^2-2a^6bc+9a^5b^3+7a^4b^4-3a^5b^2c+13a^3b^3c-9a^4b^2c^2-17a^3b^3c^2)\geq0$$

which is obviously true by Muirhead.

Answer 3

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.

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

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