# prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$

prove that $$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$$ where $$x,y,z$$ are positives such that $$xy+yz+xz=1$$

By Holder;$$\left(\sum_{cyc} \frac{1}{{(x+y)}^2} \right){\left(\sum yz+zx \right)}^2\ge {\sum \left(z^{2/3} \right)}^{3}$$.

Hence it suffices to prove $$\sum {\left(z^{2/3} \right)}^{3}\ge 3^{2/3}$$ which is falsse.

I was able to get a weaker bound.since $$xy+yz+zx=1$$ we have $$x^2+y^2\le 2$$. By C-S and $$x^2+y^2<2$$ $$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge \frac{1}{2}\sum \frac{1}{x^2+y^2}>3/4$$

I am interested on a solution using standard inequalities (C-S AM-GM,chebyshov etc) rather than fully expanding and using uvw/pqr/schur.

• See this thread on AoPS, which does use uvw but is still quite elegant. Found using Approach0. Nov 1, 2020 at 9:20
• C-S is weak in direct form I think; well what is the source of this problem? Nov 1, 2020 at 9:47
• @BookOfFlames It's Iran TST 96 Nov 1, 2020 at 9:52

## 2 Answers

This is the solution by Vo Quoc Ba Can.

Need to prove: $$(xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\geqq \frac{9}{4}$$ Wlog, assume $$x\geq y \geq z,$$ we have the following lemma: $$\sum\limits_{cyc} \frac{1}{(x+y)^2}\geqq \frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\quad \quad(1)$$ $$\Leftrightarrow \left(\frac{1}{x+z}-\frac{1}{y+z}\right)^2\geqq \frac{(x-y)^2}{4xy(x+y)^2}$$ $$\Leftrightarrow 4xy(x+y)^2 \geqq (x+z)^2(y+z)^2$$ Since $$x\geqq y \geqq z \therefore 4xy\geqq 4y^2\geqq (y+z)^2;(x+y)^2 \geqq (x+z)^2.$$

So $$(1)$$ is true, need to prove: $$(xy+yz+zx)\left(\frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\right)\geqq\frac{9}{4}$$ But it's $$(x+y)(y+z)(z+x)\geq 8xyz,$$ which is true by AM-GM.

• Do you know Chinese? I read your blog just now, that's why I'm asking. Nov 1, 2020 at 9:33
• @TobyMak I know a little, but I can't write it well. Nov 1, 2020 at 9:52

We need to prove that: $$\sum_{cyc}\frac{1}{(x+y)^2}\geq\frac{9}{4(xy+xz+yz)}$$ or $$\sum_{cyc}\left(\frac{1}{(x+y)^2}-\frac{3}{4(xy+xz+yz)}\right)\geq0$$ or $$\sum_{cyc}\frac{4xz+4yz-2xy-3x^2-3y^2}{(x+y)^2}\geq0$$ or $$\sum_{cyc}\frac{(z-x)(3x+y)-(y-z)(3y+x)}{(x+y)^2}\geq0$$ or $$\sum_{cyc}(x-y)\left(\frac{3y+z}{(y+z)^2}-\frac{3x+z}{(x+z)^2}\right)\geq0$$ or $$\sum_{cyc}(x-y)^2(3xy+xz+yz-z^2)(x+y)^2\geq0.$$ Now, let $$x\geq y\geq z$$.

Thus, $$\sum_{cyc}(x-y)^2(3xy+xz+yz-z^2)(x+y)^2\geq$$ $$\geq\sum_{cyc}(x-y)^2(xz+yz-z^2)(x+y)^2\geq$$ $$\geq(x-z)^2y(x+z-y)(x+z)^2+(y-z)^2x(y+z-x)(y+z)^2\geq$$ $$\geq(y-z)^2y(x-y)(x+z)^2+(y-z)^2x(y-x)(y+z)^2=$$ $$=(x-y)^2(y-z)^2(xy-z^2)\geq0.$$

• wow! but how did you even comeup with such an idea? Nov 1, 2020 at 11:51
• @AlbusDumbledore this is the SOS method. Nov 1, 2020 at 12:25