Show that for a, b, c > 0 with a + b + c = 1, $(a + \frac 1a)^2 + (b + \frac 1b)^2 + (c + \frac 1c)^2 ≥ \frac {100}{3}$

Question

For this question I took the function as $x^2$ and using the second derivative I found it to be 2. This is greater than 0 for all x values meaning that the function is convex. Then I used Jensen's Inequality which got me $$\frac {(a + \frac 1a)^2 + (b + \frac 1b)^2 + (c + \frac 1c)^2}{3} ≥ \frac {(a + b + c + \frac 1a + \frac 1b + \frac 1c)^2}{9}$$ I'm not sure how to find the value of $ \frac 1a + \frac 1b + \frac 1c$ or if this even is the correct approach. Can I have some hints for the next step or direction of attack?

Answer 1

Putting it all together from the comments.

$$\left(a + \frac 1a\right)^2 + \left(b + \frac 1b\right)^2 + \left(c + \frac 1c\right)^2 ≥ \frac {(a + b + c + \frac 1a + \frac 1b + \frac 1c)^2}{3}$$

$$≥ \frac {\left(a + b + c + \frac{9}{a+b+c} \right)^2}{3} = \frac{100}{9}$$

Note that applying Cauchy-Schwarz or QM-AM yields the first inequality, so no need to use Jensen's here.