How do I prove that for $ x,y,z > 0$ $\frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y} > 3/2$?

Question

I'm solving a graduate entrance examination problem. We are required to establish the inequality using the following result:

for $x,y > 0$, $\frac{x}{y} + \frac{y}{x} > 2$ (1), which is easy to prove as it is equivalent to $(x - y)^2 > 0$.

But when it comes to an inequality combining $x, y, z$, I got stuck as I've tried to develop the expression into one single fraction and obtain something irreducible.

Any hints ? My intuition tells me that for $x,y,z >0$, any fraction of the form $\frac{x}{y+z}$ is greater than 1/2. As there are three fractions of this kind with mute variables playing symmetrical roles, we get: $1/2 + 1/2 + 1/2 = 3/2$.

I just don't figure out how to play with the result (1).

Answer 1

Apply your formula three times:

$$\frac{x+y}{x+z}+\frac{x+z}{x+y}>2$$

$$\frac{y+x}{y+z}+\frac{y+z}{y+x}>2$$

$$\frac{z+x}{z+y}+\frac{z+y}{z+x}>2$$

Combining gives:

$$\frac{x+y}{x+z}+\frac{x+z}{x+y}+\frac{y+x}{y+z}+\frac{y+z}{y+x}+\frac{z+x}{z+y}+\frac{z+y}{z+x}>6$$

$$\frac{x+y+2z}{x+y}+\frac{x+2y+z}{x+z}+\frac{2x+y+z}{y+z}>6$$

$$1+\frac{2z}{x+y}+1+\frac{2y}{x+z}+1+\frac{2x}{y+z}>6$$

$$2\left(\frac{z}{x+y}+\frac{y}{x+z}+\frac{x}{y+z}\right)>3$$

$$\frac{z}{x+y}+\frac{y}{x+z}+\frac{x}{y+z}>\frac{3}{2}$$

Answer 2

Hint: put $a=x+y, b=y+z. c=z+x$. Then $2x=a-b+c, 2y=a+b-c, 2z=-a+b+c$. Rewrite the inequality in terms of $a,b,c$ and try to apply the lemma you are supposed to use.

Answer 3

Using the Rearrangement Inequality:

$x,y,z$ and $\frac1{y+z},\frac1{z+x},\frac1{x+y}$ have the same ordering, so $$(x,y,z)\left(\frac1{y+z},\frac1{z+x},\frac1{x+y}\right)\geq(x,y,z)\left(\frac1{x+y},\frac1{y+z},\frac1{z+x}\right)$$ and $$(x,y,z)\left(\frac1{y+z},\frac1{z+x},\frac1{x+y}\right)\geq(x,y,z)\left(\frac1{z+x},\frac1{x+y},\frac1{y+z}\right).$$ Adding both inequalities gives $$2(x,y,z)\left(\frac1{y+z},\frac1{z+x},\frac1{x+y}\right)\geq3.$$

Answer 4

Proof:

If $x,y,z>0$,then:$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$$$\geq3(xy+yz+zx)$$

So:$$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}$$$$=\frac{x^2}{xy+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{yz+zx}$$$$\geq\frac{(x+y+z)^2}{2(xy+yz+zx)}\geq\frac3{2}$$