# Inequality Proofs

Inequality Proof can be done in many ways. For proving $A \ge B$, one of the easiest ways is to show $A-B \ge 0$.

### Worked Example of Inequality Proofs

If $a, b$ and $c$ are positive real numbers and $a+b\ge c$, prove that $\displaystyle \frac{a}{1+a} + \frac{b}{1+b} \ge \frac{c}{1+c}$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \frac{a}{1+a} + \frac{b}{1+b}-\frac{c}{1+c} &= \frac{a(1+b)(1+c)+b(1+a)(1+c)-c(1+a)(1+b)}{(1+a)(1+b)(1+c)} \\ &= \frac{a+ab+ac+abc+b+ab+ac+abc-c-ac-bc-abc}{(1+a)(1+b)(1+c)} \\ &= \frac{(a+b-c)+2ab+abc}{(1+a)(1+b)(1+c)} \\ &\ge 0 \ \ \ \color{red} a+b-c \ge 0, \text{ and } a, b, c \text{ are positive} \\ \therefore \frac{a}{1+a} + \frac{b}{1+b} &\ge \frac{c}{1+c} \end{aligned}

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