XYZ! It's easy as abc! :D

Let $x, y, z$ be positive numbers such that:

$(1)$ $x = \frac{a}{a+b}$

$(2)$ $y = \frac{b}{b+c}$

$(3)$ $z = \frac{c}{c+a}$

Prove the following:

$x+y+z > 1$

#Algebra #NumberTheory #Inequalities #Proofs #ProofProblems

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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Comments

Hint: fund $\frac{1}{x}$ and similar, apply AM-HM on $x,y,z$ , and then try to simplify

Dinesh Chavan - 6 years, 12 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$