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 $1$?
 A: 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 $f(x)>1$ $\forall x\in (0,4).$
A: Note that $$(x-5)^2+\frac{9(4-x)}{4x}-1=\frac{(x-4)(4x^2-24x-9)}{4x}$$
A: 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.
A: 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.
