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!
Best Answer
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.