# Proof for $\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} + \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}} + \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}$? [duplicate]

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: $$$$\begin{split} \displaystyle \sum_{cyc} \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} & \geq \frac{4}{3} \\ \stackrel{(*)}{\Rightarrow} \sum_{cyc} \frac{4a^2}{(a^2+2)(b^2+2)} & \geq \frac{4}{3} \\ \Rightarrow 3 \sum_{cyc} a^2(c^2+2) & \geq (a^2+2)(b^2+2)(c^2+2) \\ \Rightarrow \sum_{cyc} 2a^2+ \sum_{cyc} a^2b^2 & \geq 72 \end{split}$$$$ 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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 6 '16 at 9:23

• Generally not valid. For example, prove that 3>5. You claim 10>3 and 10>5. – karakfa Jan 5 '16 at 21:48
• You need to show for that $x≥z$ and $z≥y$ for that $z$ – karakfa Jan 5 '16 at 22:55
• 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! – 貓貓吃狗狗 Jan 6 '16 at 2:46
• As commented above, $$\sqrt{1+k^3} \le \frac{k^2+2}2 \implies \frac1{\sqrt{1+k^3} } \ge \frac2{k^2+2}$$ – Macavity Jan 6 '16 at 9:26