Try Tackling This Radical Inequality!

Note that we must have $5x + 1>0$ and $x^2-3x+4> 0$ , otherwise the radical expressions are non-real numbers, so $x>-\dfrac15$ . It's trivial to show that $x = 0$ is a solution. What's left to do is to find the positive integer roots of $x$ satisfying the given inequality.

The inequality can be rearranged to

$\dfrac2{\sqrt{5x+1}} + \dfrac2{\sqrt{x^2-3x+4}} \geq 1 + \dfrac2{x+1} > 1 \qquad \qquad x> 0$

Or equivalently,

$\dfrac12 \left( \dfrac1{\sqrt{5x+1}} + \dfrac1{\sqrt{x^2-3x+4}} \right) > \dfrac14 \qquad \qquad x> 0$

Since both $\dfrac1{\sqrt{5x+1}}$ and $\dfrac1{\sqrt{x^2-3x+4}}$ are strictly positive numbers for $x>0$ , we can apply the Power mean inequality :

$\begin{aligned} && QM \left(\dfrac1{\sqrt{5x+1}} , \dfrac1{\sqrt{x^2-3x+4}} \right) \geq AM \left(\dfrac1{\sqrt{5x+1}} , \dfrac1{\sqrt{x^2-3x+4}} \right) > \dfrac14 \\ && \Leftrightarrow \sqrt{ \dfrac{1/(5x+1) + 1/(x^2-3x+4) }2 } > \dfrac14 \\ && \Leftrightarrow \dfrac1{5x+1} + \dfrac1{x^2-3x+ 4} > \dfrac18 \\ && \Leftrightarrow 5x^3 - 22x^2 + x - 36 < 0 \\ \end{aligned}$

Hence, the positive values that we're looking are bounded under the inequality $f(x) < 0$ , where $f(x) = 5x^3 - 22x^2 + x - 36$ . Since the cubic discriminant of this polynomial is easily computed to be a negative value, then it has precisely one real root. Because $f(4) < 0 < f(5)$ , we know by intermediate value theorem that $f(x)$ only has a real root in the interval $(4,5)$ .

In other words, the only potential positive values of $x$ that satisfy the inequality in question are $x= 1,2,3,4$ . Trial and error shows that only $x=4$ does not satisfy the original inequality. Thus, the only solutions are $x = 0,1,2,3$ .

Here we go.

First, the constrain of $x$ is $x>\frac{-1}{5}$ and due to this, $RHS>0$ . The inequality can be written as $\frac{1}{\sqrt{5x+1}}-\frac{1}{x+1}+\frac{1}{\sqrt{x^2-3x+4}}-\frac{1}{2}\geq 0$ $\Leftrightarrow\frac{x+1-\sqrt{5x+1}}{(x+1)\sqrt{5x+1}}+\frac{2-\sqrt{x^2-3x+4}}{2\sqrt{x^2-3x+4}}\geq 0$ $\Leftrightarrow\frac{x^2-3x}{(x+1)\sqrt{5x+1}(x+1+\sqrt{5x+1})}-\frac{x^2-3x}{2\sqrt{x^2-3x+4}(\sqrt{x^2-3x+4}+2)}\geq 0$ $\Leftrightarrow (x^2-3x)\bigg[\frac{2\sqrt{x^2-3x+4}(\sqrt{x^2-3x+4}+2)-(x+1)\sqrt{5x+1}(x+1+\sqrt{5x+1})}{(x+1)\sqrt{5x+1}(x+1+\sqrt{5x+1}).2\sqrt{x^2-3x+4}(\sqrt{x^2-3x+4}+2)}\bigg]\geq 0$ Since $(x+1)\sqrt{5x+1}(x+1+\sqrt{5x+1}).2\sqrt{x^2-3x+4}(\sqrt{x^2-3x+4}+2)>0; \forall x>\frac{-1}{5}$ , we can multiply both side to clear it out. It left us with $(x^2-3x)\big[2\sqrt{x^2-3x+4}(\sqrt{x^2-3x+4}+2)-(x+1)\sqrt{5x+1}(x+1+\sqrt{5x+1})\big]\geq 0$ $\Leftrightarrow (x^2-3x)\big[4\sqrt{x^2-3x+4}-(x+1)^2\sqrt{5x+1}-3x^2-12x+7\big]\geq 0$ $\Leftrightarrow (x^2-3x)\big\{4[\sqrt{x^2-3x+4}-(x+1)]+(x+1)^2(2-\sqrt{5x+1})-5x^2-12x+9\big\}\geq 0$ $\Leftrightarrow (x^2-3x)(3-5x)\bigg[\frac{4}{x+1+\sqrt{x^2-3x+4}}+\frac{(x+1)^2}{2+\sqrt{5x+1}}+x+3\bigg]\geq 0$ We see that $\frac{4}{x+1+\sqrt{x^2-3x+4}}+\frac{(x+1)^2}{2+\sqrt{5x+1}}+x+3>0;\forall x>\frac{-1}{5}$ so what we have left is $(x^2-3x)(3-5x)\geq 0$ . The inequality gives $x\in \big(\frac{-1}{5};0\big]\cup\big[\frac{3}{5};3\big]$ after solving and combining with the constrain. So there are 4 intergers $0;1;2;3$ satisfy the problem

We must have integer $x \ge 0$ for the inequality to be possible. Note that $\frac{1}{\sqrt{5x+1}}$ and $\frac{1}{\sqrt{x^2 - 3x + 4}}$ are decreasing functions of $x \ge 2$ , while $\frac{x+3}{2(x+1)}$ is a decreasing function of $x \ge 0$ tending to $\frac12$ as $x \to \infty$ . Write $f(x) \; =\; \frac{1}{\sqrt{5x+1}} + \frac{1}{\sqrt{x^2 - 3x + 4}} - \frac{x+3}{2(x+1)}$ We can calculate that $f(0) = f(3) = 0$ , while $f(1),f(2) > 0 > f(4)$ . Since $f(x) \; \le \; \frac{1}{\sqrt{26}} + \frac{1}{\sqrt{14}} - \frac12 < 0 \qquad \forall x \ge 5$ we see that $f(x) \ge 0$ precisely when $x = 0,1,2,3$ , making the answer $\boxed{4}$ .