which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$ Which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$?
Squaring both sides will give me something but I could not go any further.
 A: Hint.
$$\sqrt{a+1}-\sqrt{a}=\frac{1}{\sqrt{a+1}+\sqrt{a}}$$
A: Brute force solution, Only Squarings:
Let's assume:
$$\sqrt{7}-\sqrt{6} \geq \sqrt{6}-\sqrt{5}$$
after squaring
$$13 -2 \sqrt{42} \geq 11 - 2 \sqrt{30}$$
after simplification:
$$1 -\sqrt{42} \geq -\sqrt{30}$$
reversing:
$$\sqrt{30} \geq \sqrt{42}-1$$
Squaring 2nd time:
$$30 \geq 43 - 2 \sqrt{42}$$
Simplification:
$$2 \sqrt{42} \geq 13$$
Last squaring:
$$ 168 \geq 169$$ Contradiction, thefore
$$\sqrt{7}-\sqrt{6} < \sqrt{6}-\sqrt{5}$$
A: Consider the shape of the graph of the root function.  It is monotonously rising but getting flatter all the time.  Hence the differences between two values ($1$ apart on the x-axis) near the y-axis is greater than between two values ($1$ apart on the x-axis) further away from the y-axis.

A: $\sqrt{7} - \sqrt{6} = (\sqrt7 - \sqrt6 )\frac{\sqrt7 + \sqrt 6}{\sqrt7 + \sqrt 6} = \frac1{\sqrt7 + \sqrt 6} $
Similarily, 
$\sqrt6 - \sqrt 5 = \frac1{\sqrt6 + \sqrt 5}$
Since $ \sqrt 7 > \sqrt 5$,
$$\sqrt7 + \sqrt6 > \sqrt6 + \sqrt5$$ 
$$\implies \frac1{\sqrt7 + \sqrt6} < \frac1{ \sqrt6 + \sqrt5}$$
$$\implies \sqrt{7} - \sqrt{6} < \sqrt{6} - \sqrt{5}$$
So $ \sqrt6 - \sqrt5$ is the greater one.
A: As $(\sqrt7+\sqrt5)^2=12+2\sqrt{35}$ and $(\sqrt6+\sqrt6)^2=12+2\sqrt{36}$
$$(\sqrt7+\sqrt5)^2<(\sqrt6+\sqrt6)^2$$
$$\implies \sqrt7+\sqrt5<\sqrt6+\sqrt6\text{ as } a^2>b^2\iff a>b\text{ for }a,b>0$$
$$\implies \sqrt7-\sqrt6<\sqrt6-\sqrt5$$
A: Let $A = (\sqrt{7}-\sqrt{6}) - (\sqrt{6}-\sqrt{5} )$  
This comes to comes to $A = (\sqrt{7}+\sqrt{5})-2\sqrt{6}$
The square of the first is $12+2\sqrt{35}$, the square of the second is $12+2\sqrt{36}$
Since $35 \lt 36$, then $A \lt 0$, hence $\sqrt{6}-\sqrt{5}$ is bigger.
A: $7-6=6-5$
or, equivalently $(\sqrt7-\sqrt6)(\sqrt7+\sqrt6)=(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)$
hence, $$\frac{\sqrt7-\sqrt6}{\sqrt6-\sqrt5}=\frac{\sqrt6+\sqrt5}{\sqrt7+\sqrt6}<1 \text{  }\text{  }\text{  } (\text{since } \sqrt7>\sqrt5)$$
$\therefore \sqrt6-\sqrt5>\sqrt7-\sqrt6$
A: Consider the function $f(x) = \sqrt{x}$. Then the second derivative is
$$f''(x) = -\frac{1}{4}x^{-3/2} < 0$$
Hence this function is concave down, so we see that $\sqrt{7} - \sqrt{6}$ is the smaller of the two numbers.
A: To take your idea of squaring both sides, you can do one thing to make that more successful.  We are comparing $\sqrt 7-\sqrt 6$ with $\sqrt 6-\sqrt 5$ so 
$$ \sqrt 7-\sqrt 6\stackrel{?}{\lt\gt}\sqrt 6-\sqrt 5\\\sqrt 7 + \sqrt 5\stackrel{?}{\lt\gt}2\sqrt 6\\
12 + 2\sqrt{35} \stackrel{?}{\lt\gt}24 $$ but since $\sqrt {35} \lt 6$ the left is less.
A: Hint:
$$
\begin{align}
\frac1{\sqrt7+\sqrt6}\le\frac1{\sqrt6+\sqrt5}
\end{align}
$$
A: You can also use the inequality between the arithmetic mean and the quadratic mean: $$ \frac{x+y}{2} \leq \sqrt{\frac{x^2 + y^2}{2}}. $$ Setting $x = \sqrt{5}$ and $y = \sqrt{7}$ gives you $$\frac{\sqrt{5} + \sqrt{7}}{2} \leq \sqrt{6},$$ which is equivalent to $$ \sqrt{7} - \sqrt{6} \leq \sqrt{6} - \sqrt{5}.$$
A: $$\frac{f(5)+f(7)}{2} \lt f(\frac{5+7}{2}) $$
for a concave function $f$, and so for $f(x)=\sqrt x$. 
But even if you proceed on the way you have started you will get the result. 
$$\begin{eqnarray}
\sqrt{7}-\sqrt{6} &\lt& \sqrt{6}-\sqrt{5} &\mid& ^2 \\
13-2\,\sqrt{6}\,\sqrt{7} &\lt&  11-2\,\sqrt{5}\,\sqrt{6} &\mid& -11+2\sqrt{6}\sqrt{7} \mid \div 2   \\
1 &\lt& \sqrt{6}\,\sqrt{7}-\sqrt{5}\,\sqrt{6} & \mid& ^2 \\
1&\lt& 72-12\,\sqrt{5}\,\sqrt{7} & \mid& +12\sqrt{5}\sqrt{7} \\
12\,\sqrt{5}\,\sqrt{7}  &\lt& 71 & \mid& ^2 \\
12^2 \cdot 35 &\lt& 71^2  
\end{eqnarray}$$
All squaring operations are reversible and don't change the inequality relation because the LHS and the RHS of the inequalities are positive.
But $$12^2 \cdot 35 =12 \cdot 6 \cdot 2 \cdot 35=72 \cdot 70 = (71+1)(71-1)=71^2-1$$
So the last and therefore all inequalities are true.
A: $1-$ I will leave this part as it is; please read the comments below:
$$\lim_{n\rightarrow\infty} \sqrt{n+1}-\sqrt{n}=0$$
$2-$ let
$$f(n)=\sqrt{n+1}-\sqrt{n}$$ 
$$f^{'}(n)=\frac{1}{2}\left[\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n(n+1)}}\right]$$
We see that $f^{'}(n)<0$ for all $n$. This shows that  $f$ is monotonically decreasing.
As a result for any pair $n_1$ and $n_2$ s.t. $n_1>n_2$ we have $f(n_2)>f(n_1)$
Now take $n_1=6$ and $n_2=5$. this leads to $f(5)>f(6)\Rightarrow  \sqrt{6}-\sqrt{5}>\sqrt{7}-\sqrt{6}$
I hope this help now.
A: Here's how to figure it out without any arithmetic: 


*

*As you know, the graph of the square function grows steeper and steeper: The difference between $(n+1)^2$ and $n^2$ grows ever larger as $n$ increases. 

*The square root function is the inverse of the square function, so the difference grows ever smaller as $n$ increases. Therefore, the difference between $\sqrt{7}$ and $\sqrt{6}$  is smaller than the difference between $\sqrt{6}$ and $\sqrt{5}$.
A: $\sqrt 6−\sqrt 5$ is greater than $\sqrt 7−\sqrt 6$ because
the value of $\sqrt 6$ is 2.449, $\sqrt 5$ is 2.236
$\sqrt 7$ is 2.645
on subtracting $\sqrt 6−\sqrt 5$ you get 0.213 whereas on subtracting $\sqrt 7−\sqrt 6$ you get 0.196.
So, $\sqrt 6−\sqrt 5$ is greater.
