I have to prove this inequality. I was just looking for a solution to this problem as I have been stuck on it for a very long time.

Question

Show that if $a$ and $h$ are positive numbers with $h<a^2$ then $$\sqrt{a^2+h}-a<\frac{h}{2a}<a-\sqrt{a^2-h}$$

Answer 1

We need to prove that $$\sqrt{a^2-h}<\frac{2a^2-h}{2a}$$ and $$\sqrt{a^2+h}<\frac{h+2a^2}{2a}.$$ Both inequalities are obviously true after squaring of the both sides.

In the both cases we'll get $h^2>0$.

Answer 2

Here's a calculus-based solution . . .

For $x > 0$, let $f(x) =\sqrt{x}$.

Then $f$ is differentiable and $f'(x) = {\large{\frac{1}{2\sqrt{x}}}}$.

Note that $f'$ is decreasing.

Suppose $0 < h < a^2$.

\begin{align*} \text{Then}\;\;f(a^2) &= \sqrt{a^2} = a \qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\; \\[4pt] f'(a^2) &= \frac{1}{2\sqrt{a^2}}=\frac{1}{2a}\\[4pt] \end{align*} By the Mean Value Theorem, there exists $c \in (a^2,a^2+h)$ such that \begin{align*} &\frac{f(a^2 + h) - f(a^2)}{h} = f'(c)\\[6pt] \implies\;&\frac{f(a^2 + h) - f(a^2)}{h} < f'(a^2)&&\text{[since $f'$ is decreasing]}\\[6pt] \implies\;&\frac{\sqrt{a^2+h}-a}{h} < \frac{1}{2a}\\[6pt] \implies\;&\sqrt{a^2+h}-a < \frac{h}{2a}\\[6pt] \end{align*} Similarly, by the Mean Value Theorem, there exists $c \in (a^2-h,a^2)$ such that \begin{align*} &\frac{f(a^2-h) - f(a^2)}{-h} = f'(c)\\[6pt] \implies\;&\frac{f(a^2) - f(a^2-h)}{h} = f'(c)\\[6pt] \implies\;&\frac{f(a^2) - f(a^2-h)}{h} > f'(a^2)&&\text{[since $f'$ is decreasing]}\\[6pt] \implies\;&\frac{a-\sqrt{a^2-h}}{h} > \frac{1}{2a}\\[6pt] \implies\;&a-\sqrt{a^2-h} > \frac{h}{2a}\\[6pt] \end{align*} which completes the proof.

Answer 3

The easiest way for me is to use the AM-GM inequality: $A + B > 2\sqrt{AB}\,$ for all non-negative $A \not = B$. (https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means).

The inequality on the right is equivalent to: $$2a^2-h > 2 a\sqrt{a^2-h}$$ which is a direct consequence of the AM-GM inequality: $$ 2a^2-h = a^2 + (a^2-h) > 2\sqrt{a^2 (a^2-h)} = 2a\sqrt{a^2-h}. $$

Similarly, the inequality on the left is equivalent to: $$2a^2+h > 2 a\sqrt{a^2+h}$$ which is again a direct consequence of the AM-GM inequality: $$ 2a^2+h = a^2 + (a^2+h) > 2\sqrt{a^2 (a^2+h)} = 2a\sqrt{a^2+h}. $$

Answer 4

I think the easiest approach would be to notice that $$ \sqrt{a^2+h}-a=\frac{h}{\sqrt{a^2+h}+a},\qquad a-\sqrt{a^2-h}=\frac{h}{a+\sqrt{a^2-h}}. $$ After that you only need to prove $$ a+\sqrt{a^2-h}<2a<a+\sqrt{a^2+h} $$ which is as obvious as it can get.