Prove that the following inequality is true for all $m \in (0, 3)$

Question

Prove that the following inequality is true for all $m \in (0, 3)$ $$\sqrt {{m^2} + 1} + \sqrt {{{\left( {3 - m} \right)}^2} + 1} \le \sqrt {\frac{{2\left( {4{m^2} - 12m - 15} \right)}}{{(m - 3)m}}} - 1.$$

I have defined $$f\left( m \right) = \sqrt {{m^2} + 1} + \sqrt {{{\left( {3 - m} \right)}^2} + 1} ,g\left( m \right) = \sqrt {\frac{{2\left( {4{m^2} - 12m - 15} \right)}}{{(m - 3)m}}} - 1.$$

but they do not make the function monotonous. The derivative of the function $h(m)=f(m)-g(m)$ is quite complicated.

Another approach is to square both sides. I have tried squaring it four times, but it is still quite complicated. I would appreciate it if you could give me some ideas to prove the inequality. Thank you!

Answer 1

By Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} &\sqrt{m^2 + 1} + \sqrt{(3 - m)^2 + 1} + 1\\[6pt] \le{}& \sqrt{\left(\frac32 + \frac32 + 1\right)\left(\frac{m^2 + 1}{3/2} + \frac{(3 - m)^2 + 1}{3/2} + 1\right)}. \end{align*} It suffices to prove that $$ \left(\frac32 + \frac32 + 1\right)\left(\frac{m^2 + 1}{3/2} + \frac{(3 - m)^2 + 1}{3/2} + 1\right) \le \frac{{2\left( {4{m^2} - 12m - 15} \right)}}{{(m - 3)m}} $$ or $$\frac{(4m^2 - 12m + 10)(3 - 2m)^2}{3m(3 - m)} \ge 0$$ which is true.

We are done.

Answer 2

Partial answer.

Define

$$f(m) = \sqrt {\frac{{2\left( {4{m^2} - 12m - 15} \right)}}{{(m - 3)m}}} - 1 - \sqrt {{m^2} + 1} - \sqrt {{{\left( {3 - m} \right)}^2} + 1}$$

We want to prove $f(m)\geq 0$ for $m\in (0,3)$ to establish your inequality. Verify that $f(m)=f(3-m)$, indeed $4m^2-12m = 4m(m-3)$. So $f(m)=f(3-m)$ and $f$ is symmetric on $1.5$.

Now consider the Taylor series terms of $f$ with center $1.5$. $f$ is symetric on $1.5$, so all the terms of the expansion are even terms. So we have

$$f(m) = \left(-\sqrt{13}+\frac{8}{\sqrt{3}}-1\right)+\left(\frac{10}{9 \sqrt{3}}-\frac{8}{13 \sqrt{13}}\right) \left(m-\frac{3}{2}\right)^2+...\\ \approx 0.0132509 + 0.470823 (m-1.5)^2 + 0.208245 (m-1.5)^4 +0.0905318 (m-1.5)^6$$

Notice that the coefficients are positive and only for even powers (because $f$ is symetric on $1.5$). If indeed all coefficients are positive, then $f$ is minimized at $1.5$. Showing this is difficult because there is no clean closed form. Here is a graphical plot of the function around $m=1.5$.

Answer 3

The given inequality is $$\sqrt{m^2 + 1} + \sqrt{(3-m)^2 + 1} \leq \sqrt{\frac{2(4m^2 - 12m -15}{(m-3)m}} \ \text{where,} \ m \in [0, 3]$$

I took the following functions $$f(m) = \sqrt{m^2 + 1} + \sqrt{(3-m)^2 + 1}$$ $$\text{and,}\ g(m) = \frac{2(4m^2 - 12m -15}{(m-3)m}$$

The first derivative of $f(m)$ is $$f^{\prime}(m) = \frac{m}{\sqrt{m^2 + 1}} + \frac{(m - 3)}{\sqrt{(3 - m)^2 + 1}}$$ Then I solved for $f^{\prime}(m) \gt 0$ since it would tell you what the point of maxima or minima is for the given interval via the first derivative test.

$$f^{\prime}(m) \gt 0$$ $$\implies \frac{m}{\sqrt{m^2 + 1}} + \frac{(m - 3)}{\sqrt{(3 - m)^2 + 1}} \gt 0$$ $$\implies m\sqrt{(3-m)^2+1} + (m-3)\sqrt{m^2+1} \gt 0$$ $$\implies m\sqrt{(3-m)^2+1} \gt -(m-3)\sqrt{m^2+1}$$ $$\implies m^2 ((3-m)^2+1) \gt (m-3)^2 (m^2+1)$$ $$\implies m^4 - 6m^3 + 10m^2 > m^4 - 6m^3 + 10m^2 - 6m + 9$$ $$\implies m \gt \frac{3}{2}$$

Thus, $m = 1.5$ is a point of local minimum. And $f(0) = \sqrt{10} + 1\approx 4.162\dots$, $f(3) = \sqrt{10} + 1\approx 4.162\dots$, $f(1.5) = 2\sqrt{3.25} \approx 3.605\dots$.

$$\therefore f(m) \in [3.605\dots, 4.162\dots]\ \forall \ m \in [0, 3]$$

And doing the same for $g(m)$, we will get $$\begin{align} g^{\prime}(m) &= \frac{2m(m-3)(8m-12) - 2(2m-3)(4m^2 -12m -15)}{(m-3)^2 m^2}\\ &=\frac{(2m-3)\cdot(8m^2 - 24m - 8m^2 + 24m + 30)}{(m-3)^2 m^2}\\ &=\frac{30(2m-3)}{(m-3)^2 m^2} \end{align}$$

Then solve for $g^{\prime}(m) \gt 0$ to get $m \gt 1.5$, which will be a point of local minimum. If I define a function $$h(m) = \sqrt{g(m)} - 1$$

Then we will have $h(1.5) = 8/\sqrt{3} - 1 \approx 3.618\dots$, $h(0)$ and $h(3)$ approach positive infinity.

$$\therefore h(m) \in [3.618\dots, \infty] \ \forall \ m \in [0,3]$$

Since $f(m)$ and $h(m)$ never intersect, we can conclude that $f(m) \leq h(m) \ \forall \ m \in [0,3]$.