<s>[INST] Radical Inequality $\sqrt{2x-1}$ + $\sqrt{3x-2}$ > $\sqrt{4x-3}$ + $\sqrt{5x-4}$
I have attempted to solve this by squaring each side, resulting in 
$5x + 2\sqrt{2x-1}\sqrt{3x-2} - 3 > 9x + 2\sqrt{(4x-3)(5x-4)} - 7 $
$4 +  2\sqrt{2x-1}\sqrt{3x-2} > 4x + 2\sqrt{(4x-3)(5x-4)}$  
$1 + 1/2\sqrt{2x-1}\sqrt{3x-2} > x + 1/2\sqrt{(4x-3)(5x-4)}$ 
After this my thought was to square again, but I don't think that'd help too much - and I don't even know if this is correct anymore. 
 [/INST] Note that
$$ x > y \Leftrightarrow \sqrt{x} > \sqrt{y}.
$$
So if $x \geq 1$,
$$
4x-3 = 2x-1+2(x-1) \geq 2x-1
$$
and thus
$\sqrt{4x-3} \geq \sqrt{2x-1}.
$
Similarly $\sqrt{5x-4} \geq \sqrt{3x-2}$. Just put those together. If $x < 1$, inequalities are reversed.
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