How to solve a inequality with fractions and roots in denominator and numerator The inequality is like that:
$$
    \sqrt{\frac{3x+1}{2}}&gt;1
$$
I have no idea how should i begin with it.
Hint: The value $\sqrt a$ is greater than $1$ iff $a&gt;1$.
Square both sides since 1 is positive; then cross multiply; then solve. The question will then become like this: $3x - 3 &gt; 0$.

Now, equate $3(x-1)=0$ giving $x=1$.

Next, consider the real number line as follows:  

-           +      ( always take + from right and then - if power is odd;


---------1---------     otherwise the sign remains the same)

And since it gives us that $3x-3&gt;0$, it follows that $x &gt; 0$; Hence, we will take the positive value. And so, the answer $x$ belongs to $(1, \infty).$