0

$$\forall \ a, b \in \mathbb{R}, prove \hspace{5mm}\frac{|a+b|}{1 + |a+b|} \leqslant \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$

I decided to take different cases to solve this problem. For $a, b \geqslant 0$ and for $a, b < 0$, the inequality is obvious as $$\frac{|a+b|}{1 + |a+b|} = \frac{a}{1+|a+b|} + \frac{b}{1+|a+b|}$$ when $a, b \geqslant 0$ and $$\frac{|a+b|}{1 + |a+b|} = \frac{-a}{1+|a+b|} + \frac{-b}{1+|a+b|}$$ when $a, b < 0$

For $a \geqslant 0, b < 0$ and $|a| > |b|$, I was able to prove the inequality. However, I'm unable to prove in the case where $a \geqslant 0, b < 0$ and $|a| < |b|$.

ichigo gyuunyuu
  • 199
  • Also: http://math.stackexchange.com/q/698869/42969, http://math.stackexchange.com/q/563491/42969, http://math.stackexchange.com/q/983043/42969 and some more. – Martin R Mar 22 '17 at 17:58

2 Answers2

1

HINT: working backwards yields $$|a+b|(1+|a|)(1+|b|)\le (|a|+|b|)(1+|a+b|)$$ multiplying this out we get: $$|a+b|+|a||a+b|+|b||a+b|+|a||b||a+b|\le |a|+|b|+|a||a+b|+|b||a+b|+2|a|b|+2|a||b||a+b|$$ can you finish this?

Dr. Sonnhard Graubner
  • 95,283
1

We need to prove that $$\frac{|a|+|b|+2|ab|}{1+|a|+|b|+|ab|}\geq\frac{|a+b|}{1+|a+b|}$$ or $$|a|+|b|+2|ab|+|ab(a+b)|\geq|a+b|,$$ which is obvious because $|a|+|b|\geq|a+b|$.

Michael Rozenberg
  • 194,933