Show that $\frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c}$

Question

show that for $a,b,c \geq 0$ if $a \leq b+c$ then $\frac{a}{1+a} \leq\frac{b}{1+b} + \frac{c}{1+c}$ I can not get a good start in this problem. a hint will be appreciate it thanks

Answer 1

Hint: $\dfrac{a}{1+a} \leq \dfrac{b+c}{1+b+c}$.

Answer 2

Working backwards: multiply by $(1+a)(1+b)(1+c)$ on both sides (which doesn't affect the inequality, as $a,b,c \geq 0$) $$\begin{align*} a(1+b)(1+c) \leq b(1+a)(1+c) + c(1+a)(1+b) &\iff 0\leq b+ab+abc+bc+c\\ &+ca+cb+abc -a-ab-ac-abc\\ &\iff 0\leq(b+c-a)+2bc+abc \end{align*}$$ Which is evident (why?, think of what your hypothesis say).

Answer 3

$$\large \dfrac{a}{1 + a} \le \dfrac{b}{1 + b} + \dfrac{c}{1 + c}$$

$$\implies 1 - \dfrac{1}{1 + a} \le 2 - \left(\dfrac{1}{1 + b} + \dfrac{1}{1 + c}\right)$$

$$\implies \dfrac{1}{1 + a} + 1 \ge \dfrac{1}{1 + b} + \dfrac{1}{1 + c}$$

$$\implies \dfrac{2 + a}{1 + a} \ge \dfrac{2 + b + c}{(1 + b)(1 + c)}$$

But $a \le b + c \implies 2 + a \le 2 + b + c$

That means $$1 + a \le (1 + b)(1 + c)$$

$$\implies 1 + a \le 1 + b + c + bc$$

Again, $a \le b + c \implies 1 + a \le 1 + b + c$

That leaves us with $0 \le bc$, which is correct for $\forall b, c \in \mathbb R^+$

The equal sign occurs when $\displaystyle \left[ \begin{aligned} b = 0 \text{ and } a = c\\ c = 0 \text{ and } a = b \end{aligned} \right.$.

Answer 4

1 <= a+1, b+1, c+1 and a+1 < b+1, c+1

Therefore, we have a+1 < (1+a)(1+b) < (1+c)(1+b)

a/(1+a) * (a+1) <= {b/(b+1) + c/(c+1)} * {(1+c)*(1+b)}

we have a <= b*(1+c) + c*(1+b) = b + c + 2bc which is true, since a < b+c and bc>0.

Answer 5

$$\frac{b}{1+b} + \frac{c}{1+c} = \frac{b+2bc+c}{(1+b)(1+c)}$$

So after multiplying the $$\frac{a}{1+a} \le \frac{b+2bc+c}{(1+b)(1+c)} $$ by $(1+a)(1+b)(1+c)$ you get:

$$a(1+b)(1+c) \le (b+2bc+c)(1+a)$$

This leads to:

$$a+abc \le b+c+2bc+2abc$$

The proof is trivial. Observe that $a\le b+c$ and $2abc > abc$.

Hope it helped.