Calculus

Applications of the Derivative

Applications of Derivative Logo

Proving of Inequalities

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

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Example 1

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

Example 2

Determine which is greater: \[{100^{101}} \text{ or } {101^{100}}?\]

Example 3

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

Example 4

Prove that for \(x \gt 0\) 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 \[\sqrt x + {\frac{1}{{\sqrt x }}} \gt 2\] is true.

Example 1.

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

Solution.

Consider the function \(f\left( x \right) = \sqrt {1 + x} - {\frac{x}{2}} - 1\) 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 \(x \gt 0\). 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: \[{100^{101}} \text{ or } {101^{100}}?\]

Solution.

Consider the function \(f\left( x \right) = {\frac{{\ln x}}{x}}\) 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 \(\frac{{\ln 100}}{{100}} \gt \frac{{\ln 101}}{{101}}\) 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 \[{\frac{a}{b} + \frac{b}{a}} \ge 2\] is valid.

Solution.

We denote \({\frac{a}{b}} = x\) and consider the function \(f\left( x \right) = x + \frac{1}{x}\) 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 \(x \gt 0.\) 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 \(f\left( 1 \right) = 1 + {\frac{1}{1}} = 2\). 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.

Solution.

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 \(x \gt 0\) (and is zero at \(x = 1\)). In this case

\[{x^2} - 2\ln x - 1 \ge 0,\;\;\; \Rightarrow 1 + 2\ln x \le {x^2}.\]

Example 5.

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

Solution.

Consider the function \(f\left( x \right) = \ln x - x + 1\). It is defined for \(x \gt 0\). Its derivative is

\[f'\left( x \right) = \left( {\ln x - x + 1} \right)^\prime = \frac{1}{x} - 1.\]

The derivative is positive for \(0 \lt x \lt 1\) and negative for \(x \gt 1.\) Consequently, the function \(f(x)\) has a maximum at the point \(x = 1\) equal

\[f\left( 1 \right) = \ln 1 - 1 + 1 = 0.\]

Thus, for \(x \gt 0\) the following inequality is valid:

\[f\left( x \right) \le 0,\;\;\; \Rightarrow \ln x - x + 1 \le 0,\;\;\; \Rightarrow \ln x \le x - 1.\]

Example 6.

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

Solution.

Consider the function \(f\left( x \right) = \sqrt x + {\frac{1}{{\sqrt x }}} - 2\). Its derivative is given by

\[f'\left( x \right)= \left( {\sqrt x + \frac{1}{{\sqrt x }} - 2} \right)^\prime = \left( {{x^{1/2}} + {x^{ - 1/2}} - 2} \right)^\prime = \frac{1}{2}{x^{ - 1/2}} - \frac{1}{2}{x^{ - 3/2}} = \frac{1}{{2{x^{1/2}}}} - \frac{1}{{2{x^{3/2}}}} = \frac{{x - 1}}{{2{x^{3/2}}}} = \frac{{x - 1}}{{2x\sqrt x }}.\]

As it can be seen, the derivative is positive for \(x \gt 1.\) Therefore, the function is increasing for \(x \gt 1.\)

Since \(f\left( 1 \right) = \sqrt 1 - {\frac{1}{{\sqrt 1 }}} - 2 = 0\), the function \(f(x)\) is positive for all \(x \gt 1.\) Thus, when \(x \gt 1,\) the following inequality holds:

\[f\left( x \right) \gt 0,\;\;\; \Rightarrow \sqrt x + \frac{1}{{\sqrt x }} - 2 \gt 0,\;\;\; \Rightarrow \sqrt x + \frac{1}{{\sqrt x }} \gt 2.\]

See more problems on Page 2.

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