Let $a, b, c$ be real numbers such that $abc=8$. Prove that: $$\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} + \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}} + \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}$$

Possible Solution:

$\sqrt{1+k^3} = \sqrt{(1+k)(1-k+k^2)} \stackrel{AM-GM}{\leq} \frac{k^2-k+1+k+1}{2} = \frac{k^2+2}{2} (*)$ Thus: $$\begin{equation} \begin{split} \displaystyle \sum_{cyc} \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} & \geq \frac{4}{3} \\ \stackrel{(*)}{\Rightarrow} \sum_{cyc} \frac{4a^2}{(a^2+2)(b^2+2)} & \geq \frac{4}{3} \\ \Rightarrow 3 \sum_{cyc} a^2(c^2+2) & \geq (a^2+2)(b^2+2)(c^2+2) \\ \Rightarrow \sum_{cyc} 2a^2+ \sum_{cyc} a^2b^2 & \geq 72 \end{split} \end{equation}$$ Which is true by applying AM-GM in both sums separately. Hence proved.

The above solution states that $\displaystyle \sum_{cyc} \frac{4a^2}{(a^2+2)(b^2+2)} \geq \sum_{cyc} \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}$

So it proves that $\displaystyle \sum_{cyc} \frac{4a^2}{(a^2+2)(b^2+2)} \geq \frac{4}{3}$ instead of $\displaystyle\sum_{cyc} \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} \geq \frac{4}{3}$

Is that or the solution correct?

Generally if we are being asked to prove that $x \geq y$ and we prove that $z \geq x$, does it suffices to prove that $z \geq y$?


marked as duplicate by chenbai, Macavity inequality Jan 6 '16 at 9:23

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  • $\begingroup$ Generally not valid. For example, prove that 3>5. You claim 10>3 and 10>5. $\endgroup$ – karakfa Jan 5 '16 at 21:48
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    $\begingroup$ Yeah I know... So the solution is wrong doesn't it? $\endgroup$ – user302454 Jan 5 '16 at 22:11
  • $\begingroup$ You need to show for that $x≥z$ and $z≥y$ for that $z$ $\endgroup$ – karakfa Jan 5 '16 at 22:55
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    $\begingroup$ Well...the solution doesn't state that $$ \sum\limits_{cyc}{\frac{4a^2}{\left(a^2+2\right)\left(b^2+2\right)}} \ge \sum\limits_{cyc}{\frac{a^2}{\sqrt{\left(1+a^3\right)\left(1+b^3\right)}}} $$ Instead, it of course means the reverse one! $\endgroup$ – 貓貓吃狗狗 Jan 6 '16 at 2:46
  • $\begingroup$ As commented above, $$\sqrt{1+k^3} \le \frac{k^2+2}2 \implies \frac1{\sqrt{1+k^3} } \ge \frac2{k^2+2}$$ $\endgroup$ – Macavity Jan 6 '16 at 9:26