Problem 55
Question
Show that \begin{equation}\text { if } \quad 0 \leq a \leq b, \quad \text { then } \quad \frac{a}{1+a} \leq \frac{b}{1+b}.\end{equation}
Step-by-Step Solution
Verified Answer
Using algebraic manipulation and applying the given condition \(0 \leq a \leq b\), we were able to prove that \(\frac{a}{1+a} \leq \frac{b}{1+b}\).
1Step 1: Convert to a Simple Inequality form
The key here is to simplify the given inequality to a more manageable one without changing its truthfulness. We can start by subtracting \( \frac{a}{1+a} \) from both sides, which gives \(0 \leq \frac{b}{1+b} - \frac{a}{1+a}\).
2Step 2: Find a Common Denominator
To further simplify the inequality, it will be helpful to find a common denominator so that the two fractions on the right can be combined. We manipulate the right-hand side to have the same denominator, yielding \(0 \leq \frac{b(1+a)-a(1+b)}{(1+b)(1+a)}\). Then simplify the numerator to get \(0 \leq \frac{ba+a^2-ba-ab}{(1+b)(1+a)}\), and further simplify to \(0 \leq \frac{a^2-ab}{(1+b)(1+a)}\).
3Step 3: Apply the Given Condition
Note that the inequality condition \(0 \leq a \leq b\) implies that \(a \leq b\) or equivalently, \(b-a \geq 0\). Replacing \(ab\) by \((b-a)a\) in the numerator yields \(0 \leq \frac{a^2-(b-a)a}{(1+b)(1+a)}\), which simplifies to \(0 \leq \frac{a^2-ba+aa}{(1+b)(1+a)}\) and eventually to \(0 \leq \frac{a^2}{(1+b)(1+a)}\). Since the denominator and numerator are positive, this inequality is equivalent to \(0 \leq a^2\), which is true for all \(a\), thus proving the given inequality.
Key Concepts
Inequality SimplificationCommon Denominator in InequalitiesApplying Inequality Conditions
Inequality Simplification
When facing complex inequalities, simplifying them is crucial for easier understanding and problem-solving. Simplification typically involves reducing the inequality to its most basic form without altering its truth.
For instance, consider the inequality \[\begin{equation} \text{if } \quad 0 \leq a \leq b, \quad \text{then } \quad \frac{a}{1+a} \leq \frac{b}{1+b}.\end{equation}\]\ In this situation, we start by eliminating the fractions to focus on the relationship between variables \(a\) and \(b\). Subtracting \(\frac{a}{1+a}\) from both sides simplifies the original inequality. This kind of manipulation is like peeling an onion; we're removing layers to reveal the core relationship we're interested in: the relationship between \(a\) and \(b\).
This process leads us to a simpler inequality that is easier to deal with and sets the stage for further steps in the proof process.
For instance, consider the inequality \[\begin{equation} \text{if } \quad 0 \leq a \leq b, \quad \text{then } \quad \frac{a}{1+a} \leq \frac{b}{1+b}.\end{equation}\]\ In this situation, we start by eliminating the fractions to focus on the relationship between variables \(a\) and \(b\). Subtracting \(\frac{a}{1+a}\) from both sides simplifies the original inequality. This kind of manipulation is like peeling an onion; we're removing layers to reveal the core relationship we're interested in: the relationship between \(a\) and \(b\).
This process leads us to a simpler inequality that is easier to deal with and sets the stage for further steps in the proof process.
Common Denominator in Inequalities
Solving inequalities with fractions often involves the crucial step of finding a common denominator. This allows us to combine terms and better understand the nature of the inequality.
For example, in our exercise, we have \(0 \leq \frac{b}{1+b} - \frac{a}{1+a}\). To manage this, we seek a common denominator, which turns out to be \((1+b)(1+a)\), allowing us to merge the two fractions into a single expression. With this maneuver—akin to syncing different beats into a coherent rhythm—we've now set a stage where the inequality's behavior becomes clearer.
Getting a common denominator is a valuable tool, which facilitates the cancellation of terms and leads to a more refined form of the inequality.
For example, in our exercise, we have \(0 \leq \frac{b}{1+b} - \frac{a}{1+a}\). To manage this, we seek a common denominator, which turns out to be \((1+b)(1+a)\), allowing us to merge the two fractions into a single expression. With this maneuver—akin to syncing different beats into a coherent rhythm—we've now set a stage where the inequality's behavior becomes clearer.
Getting a common denominator is a valuable tool, which facilitates the cancellation of terms and leads to a more refined form of the inequality.
Applying Inequality Conditions
Finally, we must consider the conditions given in the problem when proving inequalities. Understanding and correctly applying these conditions are often the key to unlocking the proof.
In our exercise, we apply the condition \(0 \leq a \leq b\) to simplify the numerator of the inequality. By recognizing that \(b-a \geq 0\), we can rewrite the expression involving \(ab\) as \((b-a)a\), which helps in simplifying the inequality further. This strategic move is like applying a filter to see the essential parts of a complex picture more clearly.
Each step taken by applying the conditions provided aligns closer to proving the inequality valid. By showing that the final expression \(0 \leq \frac{a^2}{(1+b)(1+a)}\) holds true for all non-negative \(a\), we effectively confirm the original inequality's truth through these systematic applications of given conditions.
In our exercise, we apply the condition \(0 \leq a \leq b\) to simplify the numerator of the inequality. By recognizing that \(b-a \geq 0\), we can rewrite the expression involving \(ab\) as \((b-a)a\), which helps in simplifying the inequality further. This strategic move is like applying a filter to see the essential parts of a complex picture more clearly.
Each step taken by applying the conditions provided aligns closer to proving the inequality valid. By showing that the final expression \(0 \leq \frac{a^2}{(1+b)(1+a)}\) holds true for all non-negative \(a\), we effectively confirm the original inequality's truth through these systematic applications of given conditions.
Other exercises in this chapter
Problem 55
State whether the function is odd, even, or neither. $$f(x)=\frac{x}{x^{2}-9}$$
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Write the expression in factored form. \(4 x^{2}+12 x+9\).
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The points are the vertices of a triangle. State whether the triangle is isosceles (two sides of equal length). a right triangle, both of these, or neither of t
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Give the domain and range of the function. $$f(x)=2 \cos 3 x$$.
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