# If $a^2+b^2+c^2=1$ so $\sum\limits_{cyc}\frac{1}{(1-ab)^2}\leq\frac{27}{4}$

Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that: $$\frac{1}{(1-ab)^2}+\frac{1}{(1-ac)^2}+\frac{1}{(1-bc)^2}\leq\frac{27}{4}$$

This inequality is stronger than $\sum\limits_{cyc}\frac{1}{1-ab}\leq\frac{9}{2}$ with the same condition,

which we can prove by AM-GM and C-S: $$\sum\limits_{cyc}\frac{1}{1-ab}=3+\sum\limits_{cyc}\left(\frac{1}{1-ab}-1\right)=3+\sum\limits_{cyc}\frac{ab}{1-ab}\leq$$ $$\leq3+\sum\limits_{cyc}\frac{(a+b)^2}{2(2a^2+2b^2+2c^2-2ab)}\leq3+\sum\limits_{cyc}\frac{(a+b)^2}{2(a^2+b^2+2c^2)}\leq$$ $$\leq3+\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)=\frac{9}{2},$$ but for the starting inequality this idea does not work.

By the way, I have a proof of the following inequality.

Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=3$. Prove that: $$\sum\limits_{cyc}\frac{1}{(4-ab)^2}\leq\frac{1}{3}$$ (we can prove it by SOS and uvw).

This inequality is weaker and it not comforting.

We can assume of course that all variables are non-negatives.

Thank you!

Let $x = \dfrac{1}{1-ab}$, $y = \dfrac{1}{1-bc}$, $z = \dfrac{1}{1-ac}$.

We have proven that

$$x + y + z \leq \dfrac{9}{2}.$$

Therefore, $x + y + z = \dfrac{9}{2} - \varepsilon$ for some $0 \leq \varepsilon \leq \dfrac{9}{2}$. If we assume that $x,y,z \geq 0$, which we can, we can square both sides and obtain

$$(x^2 + y^2 + z^2) + (2xy + 2yz + 2xz) = \bigg{(} \dfrac{9}{2} - \varepsilon \bigg{)}^2 \leq \dfrac{81}{4}$$

We can say that $x^2 + y^2 + z^2 \leq \dfrac{81}{4} - 2(xy + yz + xz)$. We now seek to minimize $xy + yz + xz$ in order to maximize the upper bound.

We are given that $a^2 + b^2 + c^2 = 1$, and since $xy + yz + xz$ is a symmetric polynomial in $x,y,z$, we should seek $x=y=z>0$ as our minimum, which can be shown the be the case using very straightforward techniques from calculus (Second derivative test should suffice, proving the minimum for $a,b,c$ suffices). We find that $x=y=z$ when $a=b=c=\dfrac{1}{\sqrt{3}}$, so that $x=y=z=\dfrac{1}{1-1/3}=\dfrac{3}{2}$. We can then say that

$$xy + yz + xz \geq 3*(3/2)^2 = 3*(9/4) = 27/4$$

Therefore,

$$x^2 + y^2 + z^2 \leq \dfrac{81}{4} - 2*\dfrac{27}{4} = \dfrac{81}{4} - \dfrac{54}{4} = \dfrac{27}{4}$$

• I don't see how you obtain the final, desired inequality. You seem to just assert it. – zibadawa timmy Apr 9 '17 at 22:33
• This comes from the fact that the function $f(x) = x^2$ is monotonically increasing on $\mathbb{R}^+$. By that I mean that for positive reals $x,h$, $x^2 < (x+h)^2$. Since $\dfrac{9}{2} \geq \dfrac{9}{2} - \varepsilon$ and since both are positive, $\bigg{(} \dfrac{9}{2} \bigg{)}^2 \geq \bigg{(} \dfrac{9}{2} - \varepsilon \bigg{)}^2$. Does that answer your question? – Will Craig Apr 9 '17 at 22:36
• $\frac{9}{2}=81/4$, so we cannot say $(\frac{9}{2} - \varepsilon)^2\le 27/4$ – didgogns Apr 9 '17 at 23:05
• I wasn't thinking, my fault. I will see if I can correct my approach. – Will Craig Apr 9 '17 at 23:05
• @didgogns I fixed the proof. Needed some calculus to make it work, but nothing difficult. – Will Craig Apr 9 '17 at 23:18

The Buffalo Way works. (I think that Michael Rozenberg knew this.)

We only need to prove the case $$a, b, c > 0$$.

After homogenization and clearing the denominators, it suffices to prove that $$f(a, b,c)\ge 0$$ where $$f(a,b,c)$$ is a homogeneous polynomial of degree $$12$$.

Due to symmetry, assume that $$c\le b\le a$$. Let $$c = 1, \ b = 1+s, \ a = 1+s+t; \ s,t\ge 0$$. Then $$f(1+s+t, 1+s, 1)$$ is a polynomial in $$s, t$$ with non-negative coefficients. We are done.