Proving inequality with mathematical induction

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I am having trouble proving these. I cannot figure out how to get to the conclusion. Here is my attempt. The stuff in red is just side work and is not part of the proof. I always get stuck on these types of problems, can someone offer some tips on how to approach these kind of problems in general? Should I start with the right hand side or the left hand side of the inequality?

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The goal is to show that 5k+1+9<6k+1. A good place to start is to note that 6k+1 = 6*6k > 6*(5k+9) by the inductive hypothesis like your sidework does. Then it suffices to show that
5k+1+9 < 6(5k+9)
 
I think I got it. So,
5k+1 +9 < 6(5k+9)
5*5k + 9 < 6*5k + 54
5*5k < 6 * 5k since 5 < 6
5*5k + 9 < 6 * 5k + 54 since 9<54

Does this look right?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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