Problem: Let $a, b, c$ be positive real numbers such that $a + b + c = \frac{1}{a^2} +\frac{1}{b^2} +\frac{1}{c^2}$. Prove that $$\begin{align} 2(a + b + c) \geq \sqrt[3]{7a^2b+1}+\sqrt[3]{7b^2c+1}+\sqrt[3]{7c^2a+1} \end{align}$$ Source: MEMO 2013
Original proof: Using AM-GM: $$\begin{align} \sqrt[3]{7a^2b+1} = 2 \sqrt[3]{a \cdot a \cdot \left (\ \frac{7b}{8} + \frac{1}{8a^2} \right ) } \end{align} \leq \frac{2}{3} \left ( a+a+\frac{7b}{8} + \frac{1}{8a^2} \right )$$ Summing up alll three inequalities with $\sqrt[3]{7a^2b+1},\sqrt[3]{7b^2c+1},\sqrt[3]{7c^2a+1}$ we get as desired.
The first equation uses a trick where we devide the variables by 8. I tried the same thing with $1^3$ from which we got a value higher than $2(a+b+c)$. Then I tried $3^3$ and that led to an inequality which was even smaller than $2(a+b+c)$. Thus we got $<$ instead of $\leq$. To be more exact summing up the three inequalities led to $\frac{62(a+b+c)}{3\cdot 27} < 2(a+b+c)$. Why is this the case? For $a=b=c=1$ we have equality. Why is this version so strange? What happens when we use different numbers in AM-GM inequalities at tricks like these? How do we know which is truly the lowest value we can get of an expression.
Because we always need to save the case of the equality occurring.
In your first proof it happens.
In your second proof it does not happen, which says that there is a mistake in your second proof.
Another way to save the equality case:
By AM-GM twice we obtain: $$\sum_{cyc}\sqrt[3]{7a^2b+1}=\frac{1}{4}\sum_{cyc}\sqrt[3]{8a\cdot8b\left(a+\frac{7}{ab}\right)}\leq\frac{1}{12}\sum_{cyc}\left(8a+8b+a+\frac{7}{ab}\right)\leq$$
$$\leq\frac{1}{12}\sum_{cyc}\left(8a+8b+a+\frac{7}{2}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\right)=2(a+b+c).$$ Also, by Holder: $$\sum_{cyc}\sqrt[3]{7a^2b+1}=\sum_{cyc}\sqrt[3]{a^2\left(7b+\frac{1}{a^2}\right)}\leq$$ $$\leq\sqrt[3]{(a+b+c)^2\sum_{cyc}\left(7b+\frac{1}{a^2}\right)}=2(a+b+c).$$