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

Question

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.

Answer 1

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!

Answer 2

(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.