Radical Inequality

Question

$\sqrt{2x-1}$ + $\sqrt{3x-2}$ > $\sqrt{4x-3}$ + $\sqrt{5x-4}$

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.

Answer 1

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 $x\in[4/5,1)$.

Answer 2

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.