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Question

If $x$ and $y$ are real numbers, show that

$$\frac{|x+y|}{1+|x+y|}≤\frac{|x|}{1+|y|}+\frac{|y|}{1+|y|}$$

I am having difficulty in proving this equation. I don't know where to start. Your help will be highly appreciated.

Martin Sleziak
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Saisi Ole Grinder
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    @Saisi Ole Grinder I think you would write this equation: $\frac{|x+y|}{1+|x+y|} \le \frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}$, right? – TheWanderer May 12 '17 at 09:16
  • Hint: The function $t/(1+t)$ is increasing on $[0,+\infty[$ and $|x+y|\leq |x|+|y|$. – Kelenner May 12 '17 at 09:18

2 Answers2

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We know that $|x+y|\leq|x|+|y|$. You need to prove $$\frac{|x+y|}{1+|x+y|}\leq\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}$$ $$or,~|x+y|.(1+|x|)(1+|y|)\leq(1+|x+y|)(|x|.(1+|y|)+|y|.(1+|x|))$$ by cross multiplication. By expanding we get $$|x+y|\leq|x|+|y|+2|x||y|+|x||y||x+y|$$ which is obviously true since $2|x||y|+|x||y||x+y|\geq0$.

QED
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Consider the function $$ f(t)=\frac{t}{1+t} $$ where $t \ge 0$. It is immediate to see that it is an increasing function on its domain. Now, since $|x+y| \le |x|+|y|$, you will have $$ \frac{|x+y|}{1+|x+y|}\le \frac{|x|}{1+|x|+|y|}+\frac{|y|}{1+|x|+|y|} $$ and, since $\frac{1}{1+|x|+|y|} \le \frac{1}{1+|x|}$, you can conclude.

P.S. Observe that the above argument holds also for the inequality you write... in the comment, I write the classical inequality students must solve in their analysis courses!

TheWanderer
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  • as $|x+y|\leq|x|+|y|$, so $\frac{1}{1+|x+y|}\geq\frac{1}{1+|x|+|y|}$. This proof is incorrect. – QED May 12 '17 at 09:27
  • @AbishankaSaha this proof is correct. –  May 12 '17 at 09:33
  • i dont understand how? can you explain? – QED May 12 '17 at 09:34
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    @Abi the function $t/(1+t)$ is increasing and $|x+y|\le |x|+|y|$. Define $z:=|x+y|$ and $w:=|x|+|y|$. Then $z/(1+z)\le w/(1+w)$. By last we have also that $|x|+|y|\ge |x|$, hence $1/(1+|x|+|y|)\le 1/(1+|x|)$ –  May 12 '17 at 09:37
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    oh sorry...i got it – QED May 12 '17 at 09:39