If $a+b+c=3$ show $a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ I have become interested in constrained relations among simple cyclic sums involving three positive variables. By simple, I mean so simple that they are also fully symmetric. The "building blocks" of the constraints and relations I have been looking at are:
$$
\sum_{\mbox{cyc}} 1 \equiv 3 \\
\sum_{\mbox{cyc}} a \\
\sum_{\mbox{cyc}} ab \\
\sum_{\mbox{cyc}} a^2 \\
\sum_{\mbox{cyc}} 1/a \\
\sum_{\mbox{cyc}} abc \equiv 3abc \\
$$
 So an easy sample would be that $$\frac{\sum_{\mbox{cyc}} abc}{\sum_{\mbox{cyc}} a^2} \leq 1$$
The first really tough one I have encountered is:

If $a$, $b$ and $c$ are positives and $a+b+c=3$, show that:
  $$a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$

I got to this while trying to prove that if $a+b+c=3$ then $a^2+b^2+c^2 \leq 1/a+1/b+1/c$; that turns out to be untrue, but only by a little bit ($27-15\sqrt{3}\approx 1.019)$.
You might show this using BW, but I would hope to find something easier to follow. 
EDIT 
The maximum ratio is $(27-15\sqrt{3})$ and it  occurs at
$$
\left(a = \sqrt{3}, b=c= \frac{3-\sqrt{3}}{2}\right)
$$
and at the two other cyclic permutations of that point.
 A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, $u=1$ and we need to prove that
$$(27-15\sqrt3)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq a^2+b^2+c^2$$ or
$$\frac{(27-15\sqrt3)v^2}{w^3}\geq3u^2-2v^2$$ or $f(w^3)\geq0,$ where
$$f(w^3)=(27-15\sqrt3)u^3v^2-(3u^2-2v^2)w^3.$$
We see that $f$  decreases, which says that it's enough to prove our inequality
for a maximal value of $w^3$, which happens for equality case of two variables.
Since $f(w^3)\geq0$ is homogeneous, it's enough to assume $b=c=1$, which gives
$$(27-15\sqrt3)(a+2)^3\left(2+\frac{1}{a}\right)\geq27(a^2+2)$$ or
$$(a-1-\sqrt3)^2(2(9-5\sqrt3)a^2+7(12-7\sqrt3)a+4(33-19\sqrt3))\geq0,$$
which is obvious.
Done!
A: (pqr method) Let $p = a + b + c = 3, q = ab + bc + ca, r = abc$.
We need to prove that $p^2 - 2q \le (27 - 15\sqrt{3})\frac{q}{r}$
or $9 - 2q \le (27 - 15\sqrt{3})\frac{q}{r}$.
Since $(a-b)^2(b-c)^2(c-a)^2 = -4p^3r+p^2q^2+18pqr-4q^3-27r^2 \ge 0$,
we have
$r\le q - 2 + \frac{2}{9}\sqrt{3(3-q)^3}$.
It suffices to prove that
$$9 - 2q \le (27 - 15\sqrt{3})\frac{q}{q - 2 + \frac{2}{9}\sqrt{3(3-q)^3}}. \tag{1}$$
Let $q = 3 - \frac{x^2}{3}$ for $x\in [0,3)$ (note: $0 < q \le 3$). Then $\sqrt{3(3-q)} = x$. (1) becomes
$$\frac{(2x + 3 - 3\sqrt{3})^2[2x^2 + (6\sqrt{3} - 9) + 63 - 36\sqrt{3}]}{6(2x+3)(3-x)}\ge 0$$
which is true.
We are done.
