Question

Use the properties of inequalities to prove the following inequalities.

If a, b, c and dare positive numbers such that $\frac{a}{b}<\frac{c}{d}$, then $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$[Hint: Show that $\frac{ad}{b}+a<c+a\text{and}a+c<\frac{cd}{d}+c$]

Short answer

If$\frac{a}{b}<\frac{c}{d}$, then $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$

Step-by-step solution

Step 1. Apply the concept of inequalities.

An inequality looks like an equation, except that in the place of the equal sign is one of the symbols, $<,>,\le ,\ge$.

Example of an inequality is $4x+7\le 19$.

An inequality is linear if each term is constant or a multiple of the variable. To solve a linear inequality, isolate the variable on one side of the inequality sign.

Rule 7: If$A\le B\text{and}B\le C$ then$A\le C$ (Inequality is transitive).

Step 2. Simplify the inequality.

Need to prove that if$a<b\text{and}c<d,\text{then}a+c<b+d.$

As $\frac{a}{b}<\frac{c}{d}$, multiplying both sides of the inequality by d, the following is obtained:

$\frac{ad}{b}<c$

Adding a on both the sides, the following is obtained:

$\frac{ad}{b}+a<a+c-----\left(1\right)$

Also, multiplying throughout the inequality$\frac{a}{b}<\frac{c}{d}$ by b, the following is obtained:

$a<\frac{bc}{d}$

When c is added to both sides of the above inequality, the following is obtained:

$a+c<\frac{bc}{d}+c------\left(2\right)$

Step 3. Prove the inequality.

Combining the inequalities (1) and (2) by using the rule 7, the following is obtained:

$\frac{ad}{b}+a<a+c<\frac{bc}{d}+c$

Divide the above inequality by$b+d$and simplify the above inequality as follows:

localid="1646043380214" $\begin{array}{c}\frac{ad}{b}+a<a+c<\frac{bc}{d}+c\\ \Rightarrow \frac{\frac{ad}{b}+a}{b+d}<\frac{a+c}{b+d}<\frac{\frac{bc}{d}+c}{b+d}\\ \Rightarrow \frac{\frac{ad+ab}{b}}{b+d}<\frac{a+c}{b+d}<\frac{\frac{bc+cd}{d}}{b+d}\\ \Rightarrow \frac{ad+ab}{b(b+d)}<\frac{a+c}{b+d}<\frac{bc+cd}{d(b+d)}\end{array}$

Simplifying further,

$\begin{array}{c}\Rightarrow \frac{a(d+b)}{b(b+d)}<\frac{a+c}{b+d}<\frac{c(b+d)}{d(b+d)}\\ \Rightarrow \frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}\end{array}$

Hence,$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$