Question

Show that \(\sqrt {1 + x} < 1 + \frac{1}{2}x\) if \(x > 0\).

Short answer

For, \(x > 0,\sqrt {1 + x} < 1 + \frac{1}{2}x\).

Step-by-step solution

Given data

The given inequality is \(\sqrt {1 + x} < 1 + \frac{1}{2}x\).

Concept of Mean value theorem

“Let\(f\)be a function that satisfies the following hypothesis:

1.\(f\)is continuous on the closed interval\(\left( {a\;,{\rm{ }}b} \right)\).

2.\(f\)is differentiable on the open interval\(\left( {a,\;b} \right)\).

Then, there is a number\(c\)in\(\left( {a\;,{\rm{ }}b} \right)\)such that\({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently,\(f(b) - f(a) = {f^\prime }(c)(b - a)\).”

Calculation to show the give inequality

The equation \(y = 1 + \frac{1}{2}x\) is a line with slope 1 and \(y\) intercept 2 and let, \(f(x) = \sqrt {1 + x} \).

Apply Mean Value theorem with \(f(x) = \sqrt {1 + x} \) on the interval, \((0,\;a)\).

\(\begin{aligned}{c}\frac{{f(b) - f(a)}}{{b - a}} &= {f^\prime }(c)\\\frac{{\sqrt {a + 1} - \sqrt {0 + 1} }}{{a - 0}} &= \frac{1}{{2\sqrt {1 + c} }}\\\frac{{\sqrt {a + 1} - 1}}{a} &= \frac{1}{{2\sqrt {1 + c} }}\end{aligned}\)

It is known that for \(c > 0,\;\frac{1}{{2\sqrt {1 + c} }} < \frac{1}{2}\).

Therefore, the equation \(\frac{{\sqrt {a + 1} - 1}}{a} = \frac{1}{{2\sqrt {1 + c} }}\) becomes:

\(\begin{aligned}{c}\frac{{\sqrt {a + 1} - 1}}{a} < \frac{1}{2}\\\sqrt {a + 1} - 1 < \frac{1}{2}a\end{aligned}\)

Add \(1\) on both sides of the above equation.

\(\sqrt {a + 1} < \frac{1}{2}a + 1\)

Replace \(a\) by \(x\).

\(\sqrt {x + 1} < \frac{1}{2}x + 1\)

Thus, for \(x > 0,\sqrt {1 + x} < 1 + \frac{1}{2}x\).