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I am trying to show the following:

Given that $|b-c| \leq a \leq b+c$ show that: $$\frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c}$$

So far I have done the following:

$a/(1+a) \leq b/(1+b) + c/(1+c) \iff$

$1-1/(1+a) \leq 2 - 1/(1+b) -1/(1+c) \iff$

$1/(1+b) -1/(1+c) \leq 1 + 1/(1+a) \iff$

$ (2+b+c)/(1+b)(1+c) \leq (2+a)/(1+a) \iff$

$ (2+b+c)(1+a) \leq (2+a)(1+b)(1+c) \iff$

$ a \leq bc(2+a)$

I am not really sure how to proceed from here- any help or hints would be much appreciated.

Blue
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stokes
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2 Answers2

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I will assume that $a,b,c > 0$ because without such an assumption we cannot even have $1+b$ and $1+c$ in the denominator.

$f(x)=\frac x {1+x}$ is an increasing function of $x$ on $[0,\infty)$. Hence $\frac a {1+a} \leq \frac {b+c} {1+b+c} =\frac b {1+b+c}+\frac c {1+b+c} \leq \frac b {1+b} +\frac c {1+c} $.

Kavi Rama Murthy
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I suppose $b,c\ge0$. $$ \frac{b}{1 + b} + \frac{c}{1 + c} \ge \frac{a}{1 + a} $$ $$ \Rightarrow \frac{b + c + 2bc}{1 + b + c +bc} \ge \frac{a}{1 + a} $$ $$ \Rightarrow b + c + 2bc + ab +ca + 2abc\ge a + ab + ca + abc $$ $$ \Rightarrow b + c + 2bc + abc \ge a $$ Add $$ b + c \ge a \quad \textrm{and} \quad bc(a +2)\ge0 $$ And we are done.

Book Of Flames
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