By Holder inequality and AM-GM we have: \[3^4=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)\geq\left(\frac{1}{\sqrt[4]{a^3b}}+\frac{1}{\sqrt[4]{b^3c}}+\frac{1}{\sqrt[4]{c^3a}}\right)^4\] But $\frac{1}{\sqrt[4]{a^3b}}\geq\frac{\sqrt{2}}{\sqrt{a^3+b}}.$ Combining the above we have the reult.

By AM-GM, ${\sum \frac{1}{\sqrt{a^3+ b}}\leq\sum \frac{1}{\sqrt{2\sqrt{a^3 b}}}= \frac{1}{\sqrt{2}}\sum \sqrt[4]{(\frac{1}{a})^3\cdot \frac{1}{b}}}\leq\frac{1}{4\sqrt{2}}\sum(\frac{3}{a}+\frac{1}{b})= \frac{3}{\sqrt{2}}.$

This problem is at the Tuymada OLympiad and my solution is with C.B.S and use that ab+bc+ac=3abc.

sqing wrote: By AM-GM, ${\sum \frac{1}{\sqrt{a^3+ b}}\leq\sum \frac{1}{\sqrt{2\sqrt{a^3 b}}}= \frac{1}{\sqrt{2}}\sum \sqrt[4]{(\frac{1}{a})^3\cdot \frac{1}{b}}}\leq\frac{1}{4\sqrt{2}}\sum(\frac{3}{a}+\frac{1}{b})= \frac{3}{\sqrt{2}}.$ $\implies$ Let $a, b, c$ be positive real numbers . Prove the inequality \[\frac{1}{\sqrt{a^3+ b}}+\frac{1}{\sqrt{b^3 + c}}+\frac{1}{\sqrt{c^3 + a}}\leq \frac{1}{\sqrt{2}}(\frac{1}{a} +\frac{1}{b} +\frac{1}{c} ).\]

By AM-GM $ \frac{1}{\sqrt{a^3+ b}}+\frac{1}{\sqrt{b^3 + c}}+\frac{1}{\sqrt{c^3 + a}} \leq \frac{1}{\sqrt{2}\sqrt[4]{a^3 b}}+\frac{1}{\sqrt{2}\sqrt[4]{b^3 c}} +\frac{1}{\sqrt{2}\sqrt[4]{c^3 a}} $ By Rearrangemrent $ \leq \frac{1}{\sqrt{2}\sqrt[4]{a^3 a}}+\frac{1}{\sqrt{2}\sqrt[4]{b^3 b}} +\frac{1}{\sqrt{2}\sqrt[4]{c^3 c}}= \frac{1}{\sqrt{2}}(\frac{1}{a} +\frac{1}{b} +\frac{1}{c} ) =\frac{3}{\sqrt{2}}.$

ali3985 wrote: By AM-GM $ \frac{1}{\sqrt{a^3+ b}}+\frac{1}{\sqrt{b^3 + c}}+\frac{1}{\sqrt{c^3 + a}} \leq \frac{1}{\sqrt{2}\sqrt[4]{a^3 b}}+\frac{1}{\sqrt{2}\sqrt[4]{b^3 c}} +\frac{1}{\sqrt{2}\sqrt[4]{c^3 a}} $ By Rearrangemrent $ \leq \frac{1}{\sqrt{2}\sqrt[4]{a^3 a}}+\frac{1}{\sqrt{2}\sqrt[4]{b^3 b}} +\frac{1}{\sqrt{2}\sqrt[4]{c^3 c}}= \frac{1}{\sqrt{2}}(\frac{1}{a} +\frac{1}{b} +\frac{1}{c} ) =\frac{3}{\sqrt{2}}.$ See #3

Positive real numbers $a, b, c$ satisfy $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = 3.$ Prove the inequality \[\frac{1}{a^2+ b}+\frac{1}{b^2 + c}+\frac{1}{c^2 + a}\leq \frac{3}{2},\]Similar to Baltic Way 2014

socrates wrote: Positive real numbers $a, b, c$ satisfy $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = 3.$ Prove the inequality \[\frac{1}{\sqrt{a^3+ b}}+\frac{1}{\sqrt{b^3 + c}}+\frac{1}{\sqrt{c^3 + a}}\leq \frac{3}{\sqrt{2}}.\] Prove that for any positive reals $a,b,c$ satisfying the condition $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$, the following is true: $$\frac{1}{{a^3+b}}+\frac{1}{b^3+c}+\frac{1}{c^3+a}\leq \frac{3}{2}$$

By Holder, $$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)\geq \left(\frac{1}{\sqrt[4]{a^3b}}+\frac{1}{\sqrt[4]{b^3c}}+\frac{1}{\sqrt[4]{c^3a}}\right)^4.$$Hence, $$3\geq \frac{1}{\sqrt[4]{a^3b}}+\frac{1}{\sqrt[4]{b^3c}}+\frac{1}{\sqrt[4]{c^3a}}\geq \sqrt{\frac{2}{a^3+b}}+\sqrt{\frac{2}{b^3+c}}+\sqrt{\frac{2}{c^3+a}},$$where the last is true by AM-GM, we are done.