<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] Assuming that all the arguments of the square roots are non negative, hence $x\geq\frac{4}{5}$, we get by squaring the following equivalent inequality:
$$\sqrt{(4x-3)(5x-4)}-\sqrt{(2x-1)(3x-2)}<2x-2,$$
that turns into:
$$\frac{2(x-1)(7x-5)}{\sqrt{(4x-3)(5x-4)}+\sqrt{(2x-1)(3x-2)}}<2(x-1).$$
Dividing by $x-1$ (and being careful in switching the direction of the inequality when needed) this boils down to proving:
$$f(x)=\sqrt{(4x-3)(5x-4)}+\sqrt{(2x-1)(3x-2)} \lessgtr 7x-5=g(x). \tag{1}$$
$g(x)$ is just the equation of the tangent line to the graphics of $f(x)$ in $x=1$, but since the LHS is a concave function as the sum of two concave functions, the initial inequality holds for any $x\in[4/5,1)$.
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