<s>[INST] 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.
 [/INST] $$\frac{f(5)+f(7)}{2} \lt f(\frac{5+7}{2}) $$
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.
</s>