I have a lot of doubts over this solution. So its probably wrong, but I hope someone can find the mistake in it. Anyways, AM-GM gives $a^3+1\ge 2a^{3/2}$. Then our inequality is : \[ \sum_{\text{cyc}}\dfrac{1}{\sqrt{a^3+1}}\le \sum_{\text{cyc}}\frac{1}{\sqrt {2}a^{3/4}}\le \frac{3}{\sqrt{2}}\left(\frac{ \displaystyle \sum_{\text{cyc}}\frac{1}{a}}{3}\right)^{3/4}=\frac{3}{\sqrt{2}} \] The last one follows from power mean. Is it correct?? I hope it is.

It's correct!

The problem is equivalent to Positive numbers $a,\ b,\ c$ satisfy $a+b+c=3$. Prove the inequality:\[\dfrac{a\sqrt{a}}{\sqrt{a^3+1}}+\dfrac{b\sqrt{b}}{\sqrt{b^3+1}}+\dfrac{c\sqrt{c}}{\sqrt{c^3+1}}\le \dfrac{3}{\sqrt{2}} \]

sqing wrote: The problem is equivalent to Positive numbers $a,\ b,\ c$ satisfy $a+b+c=3$. Prove the inequality:\[\dfrac{a\sqrt{a}}{\sqrt{a^3+1}}+\dfrac{b\sqrt{b}}{\sqrt{b^3+1}}+\dfrac{c\sqrt{c}}{\sqrt{c^3+1}}\le \dfrac{3}{\sqrt{2}} \] $(a^3+1)(3a+1)^2\geq 32a^3\iff\dfrac{a\sqrt{a}}{\sqrt{a^3+1}}\leq \frac{3a+1}{4\sqrt{2}}\Rightarrow$ $\dfrac{a\sqrt{a}}{\sqrt{a^3+1}}+\dfrac{b\sqrt{b}}{\sqrt{b^3+1}}+\dfrac{c\sqrt{c}}{\sqrt{c^3+1}}\leq \frac{3a+1}{4\sqrt{2}}+ \frac{3b+1}{4\sqrt{2}}+ \frac{3c+1}{4\sqrt{2}}=\dfrac{3}{\sqrt{2}} .$ Aiscrim wrote: Positive numbers $a,\ b,\ c$ satisfy $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3$. Prove the inequality \[\dfrac{1}{\sqrt{a^3+1}}+\dfrac{1}{\sqrt{b^3+1}}+\dfrac{1}{\sqrt{c^3+1}}\le \dfrac{3}{\sqrt{2}}. \] (N. Alexandrov) $\dfrac{1}{\sqrt{a^3+1}}+\dfrac{1}{\sqrt{b^3+1}}+\dfrac{1}{\sqrt{c^3+1}}\le \sqrt{3\left(\dfrac{a}{a^3+1}+\dfrac{b}{b^3+1}+\dfrac{c}{c^3+1}\right)}$ $\le \sqrt{\frac{3}{2}\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)} \le \dfrac{3}{\sqrt{2}}. $

$\frac{1}{\sqrt{a^3+1}}\le \frac{\sqrt{a+1}}{a^2+1} \le \frac{\sqrt{a+1}}{2a} = \frac{2\sqrt{(a+1)2}}{4\sqrt{2}a} \le \frac{(a+1)+2}{4\sqrt{2}a} = \frac{1}{4\sqrt{2}} +\frac{3}{4\sqrt{2}a} $

We can also use \[\frac{1}{\sqrt{x^3+1}}\le\frac{(3+x)\sqrt{2}}{8x}\] which isn't hard to verify and does exactly what we want to.

I think we can apply Jensen inequality for the concave function $ \ f(\frac{1}{a}) = \frac{1}{\sqrt{a^3+1}} $ and we obtain LHS $ \le \frac{3}{\sqrt{2}} $

mihaith wrote: I think we can apply Jensen inequality for the concave function $ \ f(\frac{1}{a}) = \frac{1}{\sqrt{a^3+1}} $ and we obtain LHS $ \le \frac{3}{\sqrt{2}} $ The function $f(\frac{1}{a})=\frac{1}{\sqrt{(\frac{1}{\frac{1}{a}})^3+1}}$ is not concave,we can easily prove that $\frac{1}{\sqrt{(\frac{1}{\frac{1}{a}})^3+1}}+\frac{1}{\sqrt{(\frac{1}{\frac{1}{b}})^3+1}}>2\frac{1}{\sqrt{\frac{8}{(\frac{1}{a}+\frac{1}{b})^3}+1}}$ for $a=1$ and $b=111$.

My mistake then .

Where my mistake here? http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3545869#p3545869 Thank you!

My Solution ___________________________________________________ $$\sum_{a,b,c}{\frac{1}{\sqrt{a^3+1}}}\le\sum_{a,b,c}{\frac{1}{\sqrt{2\sqrt{a^3}}}}=\sum_{a,b,c}{\frac{1}{\sqrt{2}\cdot \sqrt[4]{ a^3}}}$$Then we must prove that $$\sum_{a,b,c}{\frac{1}{\sqrt[4]{a^3}}}\le 3$$___________________________________________________ $\sum_{a,b,c}{\frac{1}{\sqrt[4]{a^3}}}=\sum_{a,b,c}{\frac{\sqrt[4]{a}}{a}}\le \frac{a+1+1+1}{4a}$ (AM-GM)$=\frac{3}{4}+\frac{3}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=3$ We are done.

From $a^3 + 1 \ge 2a$ for positive $a$, \[\sum \frac{1}{\sqrt{a^3 +1}} \le \sum\frac{1}{\sqrt{2a}} \le \sqrt{3 \times \sum \frac{1}{2a}} = \frac{3}{\sqrt{2}}.\]

Vexation wrote: From $a^3 + 1 \ge 2a$ for positive $a$, \[\sum \frac{1}{\sqrt{a^3 +1}} \le \sum\frac{1}{\sqrt{2a}} \le \sqrt{3 \times \sum \frac{1}{2a}} = \frac{3}{\sqrt{2}}.\] Why $a^3 + 1 \ge 2a$?

It's incorrect. Check $a=0.7$.

Aiscrim wrote: Positive numbers $a,\ b,\ c$ satisfy $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3$. Prove the inequality \[\dfrac{1}{\sqrt{a^3+1}}+\dfrac{1}{\sqrt{b^3+1}}+\dfrac{1}{\sqrt{c^3+1}}\le \dfrac{3}{\sqrt{2}}. \] (N. Alexandrov) By C-S we have: $(a^3+1)(a+a^2)\ge (a+a^2)^2$$\iff$ $\sqrt{a^3+1} \ge \sqrt{a(a+1)}$ Then $$\dfrac{1}{\sqrt{a^3+1}} \le \dfrac{1}{\sqrt{a(a+1)}}=\dfrac{\sqrt{2}}{\sqrt{2a(a+1)}} \le \frac{\sqrt{2}}{2}(\frac{1}{2a}+\frac{1}{a+1})\le \frac{\sqrt{2}}{2}(\frac{1}{2a}+\frac{1}{4}+\frac{1}{4a})=\frac{1}{4\sqrt{2}}(1+\frac{3}{a})$$ Hence $$\dfrac{1}{\sqrt{a^3+1}}+\dfrac{1}{\sqrt{b^3+1}}+\dfrac{1}{\sqrt{c^3+1}} \le \frac{1}{4\sqrt{2}}(3+3(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}))=\frac{3}{\sqrt{2}}$$

Tangent Line method Let $f(x)=\frac{1}{\sqrt{\frac{1}{x^3}+1}}$,then $f(x)\leq \frac{3\sqrt{2}}{8}x+\frac{\sqrt{2}}{8}$ $\Leftrightarrow$ $(x-1)^2(9x^3+24x^2+8x+1)\geq 0$ This holds when $x> 0$.So $\sum_{cyc}f\left(\frac{1}{a}\right)\leq \frac{3\sqrt{2}}{8}\cdot 3+\frac{3\sqrt{2}}{8}=\frac{3}{\sqrt{2}}$

$\sum{\frac{1}{\sqrt{a^3+1}}} \le \sum{\frac{1}{\sqrt{2}a^{\frac{3}{4}}}} \le\frac{1}{4\sqrt{2}}\sum{(\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+1)}=\frac{12}{4\sqrt{2}}=\frac{3}{\sqrt{2}}$

Let $x=\frac{1}{a}$, $y=\frac{1}{b}$, $z=\frac{1}{c}$. Then the condition is $x+y+z=3$ and the desired inequality is $$\sum_{cyc} \sqrt{\frac{x^3}{x^3+1}} \leq \frac{3}{\sqrt{2}}$$Observe that $$\sqrt{\frac{x^3}{x^3+1}} \leq \frac{1}{\sqrt{2}}+\frac{3}{4\sqrt{2}}(x-1)$$Is equivalent to $$(x-1)^2(9x^3+24x^2+8x+1)\geq 0$$Which is obviously true. Summing cyclically yields the desired.