Calculus

Applications of the Derivative

Applications of Derivative Logo

Proving of Inequalities

Solved Problems

Example 7.

Prove the inequality \[{\frac{{b - a}}{b}} \le \ln {\frac{b}{a}} \le {\frac{{b - a}}{a}}\] provided \(0 \lt a \le b\).

Solution.

Consider the logarithmic function \(f\left( x \right) = \ln x\) and apply the Lagrange formula to it on the interval \([a, b]:\)

\[\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = f'\left( \xi \right),\;\;\xi \in \left( {a,b} \right).\]

Hence,

\[\frac{{\ln b - \ln a}}{{b - a} = \frac{1}{\xi },\;\;\;} \Rightarrow \ln \frac{b}{a} = \left( {b - a} \right)\frac{1}{\xi }.\]

The right-hand side of this equation is minimal when \(\xi = b\) and, respectively, takes a maximum at \(\xi = a.\) As a result, we get the double inequality, which was required to be proved:

\[\frac{{b - a}}{b} \le \ln \frac{b}{a} \le \frac{{b - a}}{a}.\]

Example 8.

Prove the inequality \[\left| {\sin a - \sin b} \right| \le \left| {a - b} \right|.\]

Solution.

Let the function \(f\left( x \right) = \sin x\) be defined on the interval \([a, b]\). Applying the Lagrange's mean value theorem to this function, we can write the following relationship:

\[\frac{{\sin b - \sin a}}{{b - a}} = \cos \xi ,\]

where \(\xi\) is some intermediate point in the interval \((a, b).\)

Hence, we obtain

\[\sin b - \sin a = \left( {b - a} \right)\cos \xi ,\;\;\; \Rightarrow \sin a - \sin b = \left( {a - b} \right)\cos \xi .\]

We write the last expression in terms of the absolute values:

\[\left| {\sin a - \sin b} \right| = \left| {a - b} \right|\left| {\cos \xi } \right|.\]

and note that \(\left| {\cos \xi } \right| \le 1\). Consequently,

\[\left| {\sin a - \sin b} \right| \le \left| {a - b} \right|.\]

Example 9.

Prove the inequality \[\left| {\arctan a - \arctan b} \right| \le \left| {a - b} \right|.\]

Solution.

Consider the function \(f\left( x \right) = \arctan x\) on the interval \([a, b]\). Apply the Lagrange theorem:

\[\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = f'\left( \xi \right),\]

where \(\xi \in \left( {a,b} \right)\). Then

\[\frac{{\arctan b - \arctan a}}{{b - a}} = \frac{1}{{1 + {\xi ^2}}},\;\;\; \Rightarrow \arctan b - \arctan a = \left( {b - a} \right)\frac{1}{{1 + {\xi ^2}}}.\]

After multiplying both sides of the equation by \(-1\), we obtain:

\[\arctan a - \arctan b = \left( {a - b} \right)\frac{1}{{1 + {\xi ^2}}}.\]

Solve the absolute value equation:

\[\left| {\arctan a - \arctan b} \right| = \left| {a - b} \right|\left| {\frac{1}{{1 + {\xi ^2}}}} \right|.\]

Note that the relation \({\left| {\frac{1}{{1 + {\xi ^2}}}} \right|} = {\frac{1}{{1 + {\xi ^2}}}} \le 1\) is always true. In this way,

\[\left| {\arctan a - \arctan b} \right| \le \left| {a - b} \right|.\]

Example 10.

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

Solution.

We consider the function \(f\left( x \right) = {e^x} - x - 1\) and investigate its monotonicity. The derivative can be written as

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

For \(x \lt 0,\) the derivative is negative, and for \(x \gt 0\) it is positive (Figure \(1\)). Therefore, the function \(f(x)\) decreases for \(x \lt 0\) and increases for \(x \gt 0.\)

Sign of the derivative of he function y=exp(x)-x-1.
Figure 1.

At the point \(x = 0,\) it has a maximum equal

\[f\left( 0 \right) = {e^0} - 0 - 1 = 0.\]

So the function \(f(x)\) is positive everywhere except of \(x = 0.\) From this it follows that

\[f\left( x \right) \gt 0,\;\;\; \Rightarrow {e^x} - x - 1 \gt 0,\;\;\; \Rightarrow {e^x} \gt 1 + x\;\;\left( {x \ne 0} \right).\]

Example 11.

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

Solution.

We introduce the function \(f\left( x \right) = \sin x + \tan x - 2x\) and check it for monotonicity. Find the derivative:

\[f'\left( x \right) = \left( {\sin x + \tan x - 2x} \right)^\prime = \cos x + \frac{1}{{{{\cos }^2}x}} - 2.\]

Determine the intervals, in which the derivative has a constant sign. The equation \(f'\left( x \right) = 0\) can be reduced to the cubic one:

\[f'\left( x \right) = 0,\;\;\; \Rightarrow \cos x + \frac{1}{{{{\cos }^2}x}} - 2 = 0,\;\;\; \Rightarrow \frac{{{{\cos }^3}x - 2{{\cos }^2}x + 1}}{{{{\cos }^2}x}} = 0,\;\;\; \Rightarrow \left\{ {\begin{array}{*{20}{c}} {{{\cos }^3}x - 2{{\cos }^2}x + 1 = 0}\\ {\cos x \ne 0} \end{array}} \right..\]

We make the change \(\cos x = z\) and compute the roots of the cubic equation:

\[{z^3} - 2{z^2} + 1 = 0,\;\;\; \Rightarrow {z^3} - {z^2} - {z^2} + z - z + 1 = 0,\;\;\; \Rightarrow {z^2}\left( {z - 1} \right) - z\left( {z - 1} \right) - \left( {z - 1} \right) = 0,\;\;\; \Rightarrow \left( {z - 1} \right)\left( {{z^2} - z - 1} \right) = 0,\;\;\; \Rightarrow {z_1} = 1.\]

Solving the quadratic equation \({z^2} - z - 1 = 0\), we get:

\[{z^2} - z - 1 = 0,\;\;\; \Rightarrow D = 1 + 4 = 5,\;\;\; \Rightarrow {z_{2,3}} = \frac{{1 \pm \sqrt 5 }}{2} \approx - 0.62;\;1.62\]

We plot the critical points we found on the \(z\)-axis (Figure \(2\)) and determine how the sign of the derivative changes when passing through each point.

Critical points of the function y=sin(x)+tan(x)-2x.
Figure 2.

The interval \(\left( {0,{\frac{\pi}{2}}} \right)\) for the variable \(x\) corresponds to the interval \((0, 1)\) for the variable \(z\). The variable \(z\) is positive in the last interval, i.e the function \(f(x)\) is strictly increasing in the interval \(\left( {0,{\frac{\pi }{2}} }\right).\)

Since

\[f\left( 0 \right) = \sin 0 + \tan 0 - 2 \cdot 0 = 0,\]

then the function \(f(x)\) is positive in the interval \(\left( {0,{\frac{\pi }{2}}} \right).\) Consequently,

\[\sin x + \tan x - 2x \gt 0,\;\;\; \Rightarrow \sin x + \tan x \gt 2x,\;\;x \in \left( {0,\frac{\pi }{2}} \right).\]

Example 12.

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

Solution.

Consider the function \(f\left( x \right) = {\frac{{\tan x}}{x}}\). Its derivative is written as

\[f'\left( x \right) = \left( {\frac{{\tan x}}{x}} \right)^\prime = \frac{{{\frac{1}{{{{\cos }^2}x}}} \cdot x - \tan x \cdot 1}}{{{x^2}}} = \frac{{{\frac{x}{{{{\cos }^2}x}}} - \tan x}}{{{x^2}}} = \frac{{x - \sin x\cos x}}{{{x^2}{{\cos }^2}x}}.\]

In our case, the denominator in the last expression is always positive. Investigate the sign of the numerator. For this we introduce another function \(g\left( x \right) = x - \sin x\cos x = x - \frac{1}{2}\sin 2x\) and explore it using the derivative. As a result, we have

\[g'\left( x \right) = \left( {x - \frac{1}{2}\sin 2x} \right)^\prime = 1 - \cos 2x \ge 0,\]

Given that \(g\left( 0 \right) = 0,\) we conclude that the numerator in the expression for \(f'(x)\) is also positive.

Thus, the derivative \(f'(x)\) is positive over the interval \(\left( {0,{\frac{\pi }{2}}} \right)\) and therefore the function \(f(x)\) increases in this interval. Hence for \(0 \lt {x_1} \lt {x_2} \lt {\frac{\pi }{2}},\) it follows that

\[\frac{{\tan {x_2}}}{{{x_2}}} \gt \frac{{\tan {x_1}}}{{{x_1}}},\;\;\; \Rightarrow \frac{{\tan {x_2}}}{{\tan {x_1}}} \gt \frac{{{x_2}}}{{{x_1}}}.\]
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