# Prove that $\frac 1 {x+y}+\frac 1 {y+z}+\frac 1 {z+x}\geq \frac 5 2$.

Given:

1. $x,y,z\geq0$

2. $xy+yz+zx=1$

Prove that $\displaystyle \frac 1 {x+y}+\frac 1 {y+z}+\frac 1 {z+x}\geq \frac 5 2$.

I tried using Cauchy's inequality LHS $\geq\frac 9 {2a+2b+2c}$, but failed. Please give me some ideas. Thank you.

• Did you try to use Lagrange multipliers ? Some context may be usefull – Belgi Mar 3 '13 at 13:17
• I haven't heard of Lagrange multipliers... – A. Chu Mar 3 '13 at 13:18
• Did you try expanding, assuming, WLOG x<y<z and using rearrangement inequality? It should work. – user45099 Mar 3 '13 at 13:22
• I tried, but nothing happens. – A. Chu Mar 3 '13 at 13:30

assume $z$ is the smallest. if we fix the sum $x+y$,

$\dfrac{1}{x+y} + \dfrac{1}{z+x}+\dfrac{1}{z+y} = \dfrac{1}{x+y} +\dfrac{x+y+2z}{1+z^2}$, take it as a function of $z$, you may see, it is increasing.[you may differentiate it or just plugin $z=0$ and compare]

Thus $z=0$, it will be smallest.

The rest is easy.

Update:

After we get $z=0$, the restriction on $x,y,z$ turns out to be $xy=1$, and the goal is to minimize

$\dfrac{1}{x+y}+x+y$

Since now $x=\dfrac{1}{y}$, thus

$\dfrac{1}{x+y} +x+y = \dfrac{1}{x+\dfrac{1}{x}}+x+\dfrac{1}{x}$. Take $t = x+\dfrac{1}{x}$, we know that $t\ge 2$. then the objective function is to minimize

$$t+\dfrac{1}{t}$$

for $t\ge 2$.

You can take a derivative to see that $t+1/t$ is increasing when $t\ge 1$. So minimum will be obtained at $t = 2$. which is $x=y=1$.

• When you fix the sum $x+y$, $z$ can only take one value based on the restriction $xy+yz+zx$=1; is it meaningful to take the expression as a function of $z$? – Vincent Tjeng Mar 3 '13 at 15:59
• when I fix $x+y$, $z$ still can change. I only fix the sum, the product can change. – Yimin Mar 3 '13 at 16:00
• @VincentTjeng even you don't fix the sum, you can also see when $z=0$,by taking partial derivative w.r.t $z$, the function will obtain the minimum. – Yimin Mar 3 '13 at 16:02
• @Yimin: What are the rest? Minimum of $x+y+1/(x+y)$ is 2, not 5/2 ... – A. Chu Mar 5 '13 at 14:25
• @jasoncube, you have $z=0$, then there is a restriction on $xy=1$. – Yimin Mar 5 '13 at 14:32

Well,I think you can use the famouse Iran 96 inequality as a result to solve this problem. the Iran 96 inequality

Let $x,y,z\geq 0$ we have

$$\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(x+z)^{2}}\geq \frac{9}{4(xy+yz+zx)}$$ square both side,we can rewrite the inequality into $$\sum_{cyc}{\frac{1}{(x+y)^{2}}}+2\sum_{cyc}{\frac{1}{(x+y)(x+z)}}\geq \frac{25}{4}$$ Now,Using this inequality as a known result,it's suffice to prove $$\sum_{cyc}{\frac{1}{(x+y)(x+z)}}\geq 2$$ after simple homogenous,it's $$(xy+yz+xz)(x+y+z)\geq (x+y)(y+z)(z+x)$$ Or $$xyz\geq 0$$ Which is obviously true,Equality occurs if and only if $x=y=1, z=0$ or it's permutation. The proof is complete

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$. Hence, the condition does not depend on $w^3$ and we need to prove that $$\frac{\sum\limits_{cyc}(x^2+3xy)}{\prod\limits_{cyc}(x+y)}\geq\frac{5}{2}$$ or $$\frac{9u^2+3v^2}{9uv^2-w^3}\geq\frac{5}{2}$$ or $f(w^3)\geq0$, where $f$ is an increasing function.

Thus, it's enough to prove our inequality for a minimal value of $w^3$, which happens in the following cases.

1. $y=x$, $z=\frac{1-x^2}{2x}$, where $0<x\leq1$ and we get something obvious;

2. $w^3=0$.

Let $z=0$. Hence, we need to prove that $$\frac{1}{x}+\frac{1}{y}+\frac{1}{x+y}\geq\frac{5}{2}$$ or $$x+y+\frac{1}{x+y}\geq\frac{5}{2}$$ or $$(x+y-2)(2(x+y)-1)\geq0,$$ which is true because $x+y\geq2\sqrt{xy}=2$ and we are done!