Non conditional inequality

[ins-]

Let $a$, $b$, $c$ are positive real numbers. Prove the following inequality:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{9}{a+b+c} \ge \frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}$

(Оффтоп)

It is an old and not hard inequality created by me and Leonard Giugiuc. I'm posting it here, because it might be interesting and is possible to inspire new similar problems.

[Sergic Primazon]

ins- в сообщении #1111600 писал(а):

Let $a$, $b$, $c$ are positive real numbers. Prove the following inequality:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{9}{a+b+c} \ge \frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}$

(Оффтоп)

It is an old and not hard inequality created by me and Leonard Giugiuc. I'm posting it here, because it might be interesting and is possible to inspire new similar problems.

Это просто неравенство Popoviciu:

$ f : I \rightarrow \mathbb{R} $ be convex on $I$ and let $x, y, z \in I $. Then for any positive reals p, q, r:$ $$pf(x) + qf(y) + rf(z) + (p + q + r)f\left(\frac {px + qy + rz }{p + q + r}\right) \ge $$ $$ (p + q)f\left(\frac{px + qy}{p + q}\right)+(q + r)f\left(\frac{qy + rz}{q + r}\right)+(r + p)f\left(\frac{rz + px}{r + p}\right) $$

$f(x)=\frac{1}{x}$, $p=q=r=1$

[ins-]

It is interesting that: $\frac{9}{a+b+c}\le\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ (Irish MO 1998). Maybe it is the reason to like the problem. It can be solved in at least 3-4 more ways and allows generalizations and modifications.

[Sergic Primazon]

ins- в сообщении #1111698 писал(а):

It is interesting that: $\frac{9}{a+b+c}\le\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ (Irish MO 1998). Maybe it is the reason to like the problem. It can be solved in at least 3-4 more ways and allows generalizations and modifications.

$a,b,c >0$

1.$\frac{2^2}{a+b} \le \frac{1}{a}+\frac{1}{b} $ , ...

2. $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \ge \frac{3^2}{2(a+b+c)}$( С-S, Jensen ...)

[arqady]

Можно ещё проинтегрировать очевидное $\sum\limits_{cyc}\left(x^2-2xy+\sqrt[3]{x^2y^2z^2}\right)\geq0$.

[ins-]

Probably ABC method works, too.