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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 $1$?

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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 $f(x)>1$ $\forall x\in (0,4).$

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Note that $$(x-5)^2+\frac{9(4-x)}{4x}-1=\frac{(x-4)(4x^2-24x-9)}{4x}$$

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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.

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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.

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