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?
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Actually $(x+\frac{1}{x})^2$ is convex. – atzlt Aug 17 '21 at 02:17
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Sorry I made a typo. Thanks the catching it! – Pink Fluffy Unicorn Aug 17 '21 at 02:18
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1no I didn't see any typo, but it seems that you are considering the function $f(x)=x^2$. Considering another function $g(x)=(x+\frac{1}{x})^2$ can solve this question: $(\sum_{cyc}(a+1/a)^2)/3\geq((\sum_{cyc}a)/3+1/((\sum_{cyc}a)/3))^2$. – atzlt Aug 17 '21 at 02:22
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2Or you may use the AM-GM inequality $(a+b+c)\left(\frac{1}{a}+\frac{1}{b} + \frac{1}{c}\right) \ge 9$ in your lower bound .. and use the fact that $a + b + c = 1$. – r9m Aug 17 '21 at 02:56
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Oh right, Cauchy-Schwartz – Pink Fluffy Unicorn Aug 17 '21 at 03:25
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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.
Toby Mak
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