# 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.

## 1 Answer

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!

• 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 2:51
• 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