Calculus

Applications of the Derivative

Proving of Inequalities

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Problems 1-4
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Problems 5-12

In this section we consider how the derivatives can be used to prove mathematical inequalities. The general approach is to study the properties of functions in the inequality using derivatives. The most important here are the properties of monotonicity and boundedness of functions. In addition, Lagrange’s mean value theorem is often used for solving inequalities. Typical examples on this topic are listed below.

Solved Problems

Click on problem description to see solution.

 Example 1

Prove the inequality \(\sqrt {1 + x} \le 1 + \large\frac{x}{2}\) for \(x \gt 1\).

 Example 2

Determine which is greater: \({100^{101}}\) or \({101^{100}}\)?

 Example 3

Prove that for any positive numbers \(a\) and \(b\) the inequality \({\large\frac{a}{b} + \frac{b}{a} \normalsize} \ge 2\) is valid.

 Example 4

Prove that for \(x > 0\) the inequality \(1 + 2\ln x \le {x^2}\) is true.

 Example 5

Prove that for \(x \gt 0\) the inequality \(\ln x \le x – 1\) is true.

 Example 6

Show that for \(x \gt 1\) the inequality \(\sqrt x + {\large\frac{1}{{\sqrt x }}\normalsize} \gt 2\) is true.

 Example 7

Prove the inequality \({\large\frac{{b – a}}{b}\normalsize} \le \ln {\large\frac{b}{a}\normalsize} \le {\large\frac{{b – a}}{a}\normalsize}\) provided \(0 \lt a \le b\).

 Example 8

Prove the inequality \(\left| {\sin a – \sin b} \right| \le \left| {a – b} \right|\).

 Example 9

Prove the inequality \(\left| {\arctan a – \arctan b} \right| \le \left| {a – b} \right|\).

 Example 10

Prove that for \(x \ne 0\) the inequality \({e^x} \gt 1 + x\) holds.

 Example 11

Prove that the inequality \(\sin x + \tan x \gt 2x\) holds in the interval \(\left( {0,{\large\frac{\pi }{2}}} \right).\)

 Example 12

Prove the inequality \({\large\frac{{\tan {x_2}}}{{\tan {x_1}}}\normalsize} \gt {\large\frac{{{x_2}}}{{{x_1}}}}\) provided \(0 \lt {x_1} \lt {x_2} \lt \large\frac{\pi }{2}\).

Example 1.

Prove the inequality \(\sqrt {1 + x} \le 1 + {\large\frac{x}{2}\normalsize}\) for \(x \gt 1\).

Consider the function \(f\left( x \right) = \sqrt {1 + x} – {\large\frac{x}{2} \normalsize} – 1\) and find its derivative:

\[
{f’\left( x \right) = {\left( {\sqrt {1 + x} – \frac{x}{2} – 1} \right)^\prime } }
= {\frac{1}{{2\sqrt {1 + x} }} – \frac{1}{2} }
= {\frac{{1 – \sqrt {1 + x} }}{2} \le 0.}
\]

We take into account that \(f\left( 0 \right) = 1 – 0 – 1 = 0\). Therefore, \(f\left( x \right) \le 0\) for \(x > 0\). Then

\[
{\sqrt {1 + x} – \frac{x}{2} – 1 \le 0,\;\;\;\;\;}\Rightarrow
{\sqrt {1 + x} \le 1 + \frac{x}{2}.}
\]

Example 2.

Determine which is greater: \({100^{101}}\) or \({101^{100}}\)?

Consider the function \(f\left( x \right) = {\large\frac{{\ln x}}{x}\normalsize}\) and calculate its derivative:

\[f’\left( x \right)
= {{\left( {\frac{{\ln x}}{x}} \right)^\prime } }
= {\frac{{{{\left( {\ln x} \right)}^\prime } \cdot x – \ln x \cdot x’}}{{{x^2}}} }
= {\frac{{\frac{1}{x} \cdot x – \ln x}}{{{x^2}}} = \frac{{1 – \ln x}}{{{x^2}}}.}
\]

As it can be seen, the derivative is negative provided \(x \gt e.\) Then for \(x \gt e,\) the function \(f(x)\) is decreasing and, therefore, the relation \(\large\frac{{\ln 100}}{{100}} \gt \frac{{\ln 101}}{{101}}\) is true. It follows from here that

\[{101\ln 100 \gt 100\ln 101,\;\;\;\;\;}\Rightarrow
{ {100^{101}} \gt {101^{100}}.}
\]

Example 3.

Prove that for any positive numbers \(a\) and \(b\) the inequality \({\large\frac{a}{b} + \frac{b}{a} \normalsize}\ge 2\) is valid.

We denote \({\large\frac{a}{b}\normalsize} = x\) and consider the function \(f\left( x \right) = x + \large\frac{1}{x}\) given that \(x \gt 0\). Determine the points of local extrema of this function:

\[f’\left( x \right)
= {{\left( {x + \frac{1}{x}} \right)^\prime } }
= {1 – \frac{1}{{{x^2}}} = 0,\;\;\;\;\;}\Rightarrow
{{x^2} = 1,\;\;\;\;\;}\Rightarrow
{x = \pm 1.}
\]

Only one point \(x = 1\) satisfies the condition \(x > 0.\) Since the derivative changes sign from minus to plus when passing through this point (from left to right), then the point \(x = 1\) is a minimum.

The value of the function at this point is equal to \(f\left( 1 \right) = 1 + {\large\frac{1}{1}\normalsize} = 2\). Consequently,

\[{f\left( x \right) \ge 2,\;\;\;\;\;}\Rightarrow
{x + \frac{1}{x} \ge 2,\;\;\;\;\;}\Rightarrow
{\frac{a}{b} + \frac{b}{a} \ge 2.}
\]

Example 4.

Prove that for \(x \gt 0\) the inequality \(1 + 2\ln x \le {x^2}\) is true.

We introduce the function \(f\left( x \right) = {x^2} – 2\ln x – 1\). Find the critical points:

\[f’\left( x \right)
= {{\left( {{x^2} – 2\ln x – 1} \right)^\prime } }
= {2x – \frac{2}{x} = 0,\;\;\;\;\;}\Rightarrow
{\frac{{2{x^2} – 2}}{x} = 0,\;\;\;\;\;}\Rightarrow
{2{x^2} – 2 = 0,\;\;\;\;\;}\Rightarrow
{{x^2} = 1,\;\;\;\;\;}\Rightarrow
{x = \pm 1.}
\]

Of the three critical points \(x = -1\), \(x = 0\), \(x = 1,\) only the last point \(x = 1\) satisfies the condition \(x \gt 0.\) The derivative is negative to the left of this point and positive to the right. Hence, the function has a minimum at this point that is equal

\[f\left( 1 \right) = 1 – 2\ln 1 – 1 = 0.\]

Thus, \(f(x) \ge 0\) for \(x > 0\) (and is zero at \(x = 1\)). In this case

\[{{x^2} – 2\ln x – 1 \ge 0,\;\;\;\;\;} \Rightarrow {1 + 2\ln x \le {x^2}.}\]
Page 1
Problems 1-4
Page 2
Problems 5-12