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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.

Michael Rozenberg
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Mark Fischler
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2 Answers2

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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!

Michael Rozenberg
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  • I don't see how you got from $f(w^3)\geq 0$ to your next equation. If I take $b=c=1$ and write $f(w^3)\geq 0$ I get $$ (27-15\sqrt{3})(a+2)^3\left(2+\frac1a\right) \geq 27a^2(a^2-2a+2) \geq 27a^2$$ – Mark Fischler Aug 15 '17 at 19:22
  • @Mark Fischler I substituted $b=c=1$ in $(27-15\sqrt3)(a+b+c)^3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq27(a^2+b^2+c^2)$. I just like to make a homogenization before. – Michael Rozenberg Aug 15 '17 at 19:25
  • One last quibble: You want the maximum value of $w$, which you argue happens for equality of two variables. But under the actual constraint $w$ is maximized at $a=b=c=1$ and under no constraints $w$ has no maximum. So how is it obvious that you can restrict yourself to the case of two variables being equal? – Mark Fischler Aug 15 '17 at 22:20
  • @Mark Fischler For homogeneous inequality the condition is not relevant already,. We can assume $b=c$, we can assume $b=c=1000$, we can assume $b=c=1$, but the last gives an easiest inequality. If you still don't like this thing then there is possibility to assume $b=c$ and $a=3-2c$, where $0<c<1.5$, but the work after this substitution is harder, I think. – Michael Rozenberg Aug 16 '17 at 02:51
  • Yes, I have no problem with the value assigned to $b$ and $c$, just that you assume $w$ is maximized when $b=c$. After all, you are looking to maximize $w$ for fixed $u$ and $v$, and it is non-trivial that for fixed sum and sum of pair products, the product of three variables is maximized when two of the variables are equal. – Mark Fischler Aug 16 '17 at 15:40
  • @Mark Fischler If you wish I can explain this point: $a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ or $w^3=x^3-3ux^2+3v^2x$, which says that the line $y=w^3$ and the graph of $y=x^3-3ux^2+3v^2x$ have three common points. Thus, $w^3$ gets a maximal value, when the line $y=w^3$ is a tangent to the graph of $y=x^3-3ux^2+3v^2x$, which happens for equality case of two variables. – Michael Rozenberg Aug 16 '17 at 15:46
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(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.

River Li
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