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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)} \\ […]
Inequalities worked in the same way, except there was 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 & \log_{10}{0.5}=-0.3<0 \\ \hline\log_{2}{0.1}=-3.3<0 & \log_{5}{0.1}=-1.4<0 & […]
Mathematical Induction Inequality is being used for proving inequalities. It is quite often applied for subtraction and/or greatness, using the assumption in 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: Show […]
In 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 […]
Solving Inequality with Variables in the Denominator requires special care 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} \text{This is TRUE […]
Worked Example Prove that \( (2n)! > 2^n (n!)^2 \) using mathematical induction for \(n \ge 2 \). Step 1: Show it is true for \( n =2 \).\( \begin{aligned} \require{AMSsymbols} \require{color}\text{LHS } &= (2 \times 2)! = 16 \\\text{RHS } &= 2^2 \times (2!) = 8 \\\text{LHS } &> { RHS} \\\end{aligned} \)\( \therefore […]
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} […]
Absolute Value Inequalities are usually proved by the absolute value being greater than or equal to a certain value. 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{AMSsymbols} \require{color}|a| &\ge a \text{ and } |b| \ge […]
The 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 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 } a + […]
Usually, mathematical induction inequality proof requires one initial value, but in some cases, two initials are required, such as in the Fibonacci sequence. In this case, it is necessary 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 third […]