For handling inequalities involving absolute values and surds, it is required to ensure the domains before solving inequalities. the final solutions must fit in the domains. Example 1 Solve for \( x \), \( |x| \gt \sqrt{x+2} \). \( \begin{align} x+2 &\ge 0 &\color{green}{\text{domain of } \sqrt{x+2} } \\ x &\ge -2 \color{green}{\cdots (1)} \\ […]

# Tag Archives: Inequality

# Exponential Inequalities using Logarithms

Inequalities are worked in exactly the same way except that there is a change of sign when dividing or multiplying both sides of the inequality by a negative number. \begin{array}{|c|c|c|} \hline \log_{2}{3}=1.6>0 & \log_{5}{3}=0.7>0 & \log_{10}{3}=0.5>0 \\ \hline \log_{2}{2}=1>0 & \log_{5}{2}=0.4>0 & \log_{10}{2}=0.3>0 \\ \hline \log_{2}{1}=0 & \log_{5}{1}=0 & \log_{10}{1}=0 \\ \hline \log_{2}{0.5}=-1<0 & \log_{5}{0.5}=-0.4<0 […]

# Mathematical Induction Inequality

Mathematical Induction Inequality is being used for proving inequalities. It is quite often applied for the subtraction and/or greatness, using the assumption at step 2. Let’s take a look at the following hand-picked examples. Basic Mathematical Induction Inequality Prove \( 4^{n-1} \gt n^2 \) for \( n \ge 3 \) by mathematical induction. Step 1: […]

# Logarithmic Inequalities

Solving logarithmic inequalities, it is important to understand the direction of the inequality changes if the base of the logarithms is less than 1.$$\log_{2}{x} \lt \log_{2}{y}, \text{ then } x \lt y \\\log_{0.5}{x} \lt \log_{0.5}{y}, \text{ then } x \gt y \\$$Also the domain of the logarithm is positive.$$\log_{10}{(x-2)}, \text{ then } x-2 \gt 0$$ […]

# Inequality with Variables in Denominator

Solving Inequality with Variables in the Denominator requires special cares due to the direction of the inequalities. Let’s have a look at the following key points. Key Point 1 \( \begin{aligned} \displaystyle \require{color} \frac{1}{x} &\ge 2 \\ \frac{1}{x} \times x &\ge 2 \times x &\color{green} \text{Many of you may think this is TRUE.} \\ &&\color{green} […]

# Mathematical Induction Inequality Proof with Factorials

Worked Example Prove that \( (2n)! > 2^n (n!)^2 \) using mathematical induction for \(n \ge 2 \). Step 1: Show it is true for \( n =2 \).\( \begin{aligned} \require{color}\text{LHS } &= (2 \times 2)! = 16 \\\text{RHS } &= 2^2 \times (2!) = 8 \\\text{LHS } &> { RHS} \\\end{aligned} \)\( \therefore \text{It […]

# Inequality Proofs

Inequality Proofs can be done 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 \( \displaystyle \frac{a}{1+a} + \frac{b}{1+b} \ge \frac{c}{1+c} \). […]

# Absolute Value Inequalities

Absolute Value Inequalities are usually proved by the absolute value of a certain value is greater than or equal to it. The square of the value is equal to the square of its absolute value. Proof of Absolute Value Inequalities Prove \(|a| + |b| \ge |a+b|\). \( \begin{aligned} \require{color} |a| &\ge a \text{ and } […]

# Inequalities using Arithmetic Mean Geometric Mean

Arithmetic Mean of \(a\) and \(b\) is always greater than or equal to the Geometric Mean of \(a\) and \(b\), for all positive real numbers with with equality if and only if \(a = b\). This is also called AM-GM (Arithmetic Mean Geometric Mean) inequality. \(\require{color}\) $$ \begin{aligned} \frac{a + b}{2} \ge \sqrt{ab} \text{ or […]

# Mathematical Induction Inequality Proof with Two Initials

Usually, mathematical induction inequality proof requires one initial value, but in some cases, two initials are to be required, such as the Fibonacci sequence. In this case, it is required to show two initials are working as the first step of the mathematical induction inequality proof, and two assumptions are to be placed for the […]