Let
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
,
![b](/media/m/e/e/c/eec0d7323095a1f2101fc1a74d069df6.png)
,
![c](/media/m/e/a/3/ea344283b6fa26e4a02989dd1fb52a51.png)
be positive real numbers such that
![\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c](/media/m/6/f/b/6fb15ef26dfc9ab9fc51aa14d9dd4d0c.png)
. Prove that:
![\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}](/media/m/1/3/5/135394ef83cf891360f2c300a5747a5c.png)
Proposed by Juhan Aru, Estonia
%V0
Let $a$, $b$, $c$ be positive real numbers such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c$. Prove that:
$$\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}$$
Proposed by Juhan Aru, Estonia