Inequalities with radicals I have a question regarding the following inequality; It may sound petty but yet; 
The following inequality $$  \sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$$ has the solution $\frac {-5}{4}\le x \lt 1  $
During the process of solving it we need to check the correctness of the solution to $2 \sqrt{10-6x} \sqrt{2-x} \gt 11x-7$ which is: $\frac {-5}{4}\le x \le \frac {7}{11}  $ and is created due to squaring . Should I take a value of it and place it in  $\sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$  or in $2 \sqrt{10-6x} \sqrt{2-x} \gt 11x-7$ to check it's validity? To which inequality should it be applied? Or are they both equivalent in regard to this? The thing is squaring gives us more solutions than there are initially, so as I suppose what is true for the second inequality may not be true for the first..
Any advice? 

As a follow-up to my comment above, I'd like to walk through a solution of this inequality.  
$$ \sqrt{-6x+10} + \sqrt{-x + 2} > \sqrt{4x+5}$$


*

*First determine the restrictions on $x$ implied by the radicals.


$$
-6x + 10 \geq 0 \;\Rightarrow\; x \leq 5/3.
$$
$$
-x + 2 \geq 0 \;\Rightarrow\; x \leq 2.
$$
$$
4x+5 \geq 0 \;\Rightarrow\; x \geq -5/4.
$$
On these restrictions alone, we know that $-5/4 \leq x \leq 5/3$.
2 .  Use algebra to solve the inequality under the assumption that $-5/4 \leq x \leq 5/3$ (so that all radicands are guaranteed nonnegative).
$$
\begin{eqnarray*}
(\sqrt{-6x+10} + \sqrt{-x + 2})^2 &>& 4x+5 \\
-7x+12 + 2\sqrt{(-6x+10)(-x+2)} &>& 4x + 5\\
2\sqrt{(-6x+10)(-x+2)} &>& 11x - 7\\
\end{eqnarray*}
$$
Now at this point there are two cases to consider.  If $11x - 7 < 0$, then the inequality is automatically satisfied.  Therefore, we have a partial solution of $-5/4 \leq x \leq 7/11$.  On the other hand, if $11x - 7 \geq 0$, then there is still work to do.  Assume now that $11x - 7 \geq 0$ and square again:
$$
\begin{eqnarray*}
4(-6x+10)(-x+2) &>& (11x - 7)^2\\
24x^2 -88x + 80 &>& 121x^2 - 154x + 49\\
0 &>& 97x^2 - 66x - 31\\
0 &>& (x-1)(97x+31)
\end{eqnarray*}
$$
The roots of the quadratic are $x=1$ and $x=-31/97$, and since the parabola is opening upward, the last inequality is satisfied for $-31/97 < x < 1$.  Together with the previous restrictions and previous work done, we finally have a complete solution:
$$ -5/4 \leq x < 1 $$
