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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