Get the answer to "Inequality with x,y,z fractionsxy+yz+x+zx≥2"

High School
Algebra
Algebra I
Quadratic function and equation

Porter Mccullough

Answered question

2022-04-22

Inequality with x,y,z fractions

\frac{x}{y} + \frac{y}{z + x} + \frac{z}{x} \geq 2

Answer & Explanation

drenkttj9

Beginner2022-04-23Added 20 answers

Step 1
This might be pretty cumbersome to prove, if you don't see the trick is to sum 1 in both sides:
$\frac{x}{y} + \frac{y}{z + x} + \frac{z}{x} + 1 \geq 3$
And this follows directly from AM-GM:
$\frac{x}{y} + \frac{y}{z + x} + (\frac{z}{x} + 1)$
$= \frac{x}{y} + \frac{y}{z + x} + \frac{z + x}{x}$
$\geq 3 \sqrt[3]{\frac{x}{y} \cdot \frac{y}{z + x} \cdot \frac{z + x}{x}} = 3$
with equality when $x = y$ and $z = 0$ .

Annie Levine

Beginner2022-04-24Added 15 answers

Step 1
By C-S and AM-GM

\frac{x}{y} + \frac{y}{x + z} + \frac{z}{x}

= \frac{x^{2}}{x y} + \frac{y^{2}}{x y + y z} + \frac{z^{2}}{x z}

\geq \frac{{(x + y + z)}^{2}}{x y + x y + y z + x z} =

= \frac{x^{2} + y^{2} + z^{2} + 2 x y + 2 x z + 2 y z}{2 x y + x z + y z}

\geq \frac{4 x y + 2 x z + 2 y z}{2 x y + x z + y z} = 2 .

Also, we can end your work
Indeed, we need to prove that:

y z^{2} + (x^{2} - x y) z + x {(x - y)}^{2} \geq 0

which is true for

x \geq y

But for

x < y

it's enough to prove that:

x^{2} {(x - y)}^{2} - 4 x y {(x - y)}^{2} \leq 0

x {(x - y)}^{2} (x - 4 y) \leq 0,

which is obvious.

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