Inequality Proofs

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 \( \displaystyle \frac{a}{1+a} + \frac{b}{1+b} \ge \frac{c}{1+c} \).

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

Unlock your full learning potential—download our expertly crafted slide files for free and transform your self-study sessions!

Discover more enlightening videos by visiting our YouTube channel!


Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Double-Angle Formula Equation Exponent Exponential Function Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Mathematical Induction Measurement Perfect Square Perimeter Prime Factorisation Probability Product Rule Proof Pythagoras Theorem Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume

Related Articles


Your email address will not be published. Required fields are marked *