# Show that $(x-5)^2+\frac{9(4-x)}{4x}>1$ on the interval $(0,4)$

Consider the function $$f$$ given by $$f(x)=(x-5)^2+\frac{9(4-x)}{4x},$$ for $$x\in (0,4)$$.

I'm asked to show that $$f(x)>1$$ on the interval $$(0,4)$$.

I've started by recognising that $$(x-5)^2>0$$ on this interval, in which case $$f(x)>\frac{9(4-x)}{4x}.$$

How can I now show that this is greater than $$1$$?

In the interval $$(0,4)$$, $$4x>0$$ and $$9(4-x)>0$$. So $$\frac{9(4-x)}{4x}>0$$ on $$(0,4)$$.
And $$(x-5)^2>1$$ for $$\forall x\in (0,4)$$ so $$f(x)>1$$ $$\forall x\in (0,4).$$
Note that $$(x-5)^2+\frac{9(4-x)}{4x}-1=\frac{(x-4)(4x^2-24x-9)}{4x}$$
We need to prove that $$(x-6)(x-4)+\frac{9(4-x)}{4x}>0$$ or $$6-x+\frac{9}{4x}>0,$$ which is obvious.
To prove the stated condition, it is enough to prove its equivalent i.e. $$\begin{equation} 4x^3-40x^2+87x+36>0 \end{equation}$$ To find if the above equation is an increasing function i.e. $$>0$$, we find its critical points. Its critical points are: $$\begin{equation} \frac{20\pm\sqrt{139}}{6} \end{equation}$$ Thus, this leads to two intervals viz. $$I=\overbrace{\Big(0, \frac{20-\sqrt{139}}{6}\Big]}^{I_1} \cup \overbrace{\Big[\frac{20-\sqrt{139}}{6},4\Big)}^{I_2}$$ since $$4<\frac{20+\sqrt{139}}{6}$$ (enough for the proof). Now, substituting the extremities of $$I_1$$, we get $$90.39947522$$ for $$x=\frac{20-\sqrt{139}}{6}$$ and $$36$$ for $$x=0$$. Hence, the above reduced equation is strictly increasing in the interval $$I_1$$. In the interval $$I_2$$, we get, $$0$$ for $$x=4$$. Hence the function is a decreasing function in the interval $$I_2$$. But since $$4$$ is not included in the interval, it is a guarantee that the function is always strictly greater than zero. Hence proved.