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

Question

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.

Answer 1

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.

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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 $x=y=1$.

Answer 2

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