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Problem Basically the method applied is the following, we fix $a=\frac{a_1+a_2+...+a_n}{n}$, if: $$f(x)\ge f(a)+f'(a)(x-a) $$This inequality holds for all x, then summing up the inequality will give us the desired conclusion.

That is basically what is stated before the example, now, I've got quite some questions!

1) Why do we have $a=\frac{a_1+a_2+...+a_n}{n}$? why is this relevant?

2) When he applies this "tangent line trick" he clearly chooses $a=1$, is there a reason for this particular value? (I mean, there must be a reason since one has to bash through a lot of algebra, no one would do that if they didn't have a good reason to do so...)

3) The last inequality holds for $a \ge- \frac{1}{2}$, not all a's, so why does this still satisfy our initial method's constraint of 'all x'?

Edit: OP forgot to mention that variables are positive

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We see that equality is if $a=b=c=1$, so write a tangent on $$f(x) = {(3-2x)^2\over x^2+(3-x)^2}$$ at $x_0=1$ and $y_0= f(1) = 1/5$. This line has equation $$y = {1\over 5}- {18\over 25}(x-1)$$

Now check if $f(x)\geq y$ for all $x\in (0,3)$ and thus a conclusion.

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