Observe that the function $f(x)=\frac{1}{\sqrt{x^3}}$ is convex for $x>0$ (as $f''(x)=\frac{15}{4}\cdot x^{-\frac{7}{2}}>0$ since $x>0$). We rewrite the given inequality as $$\frac{a}{a+b+c}\cdot\left(\frac{1}{\sqrt{(a+2b)^3}}\right)+\frac{b}{a+b+c}\cdot\left(\frac{1}{\sqrt{(b+2c)^3}}\right)+\frac{c}{a+b+c}\cdot\left(\frac{1}{\sqrt{(c+2a)^2}}\right)\geq\frac{1}{\sqrt{(a+b+c)^3}}$$and apply the (weighted form of) Jensen's Inequality, obtaining $$LHS\geq\sqrt{\frac{1}{\left(\frac{a(a+2b)+b(b+2c)+c(c+2a)}{a+b+c}\right)^3}}=\frac{1}{\sqrt{(a+b+c)^3}}$$as needed. Equality is reached if and only if $a=b=c$.

Let $a, b, c$ are positive real numbers. Prove that $$\frac{1}{\sqrt{a + 2b}}+\frac{1}{\sqrt{b + 2c}} +\frac{1} {\sqrt{c + 2a}} \ge \frac{3}{\sqrt{a + b + c}}$$

parmenides51 wrote: If $a, b, c$ are positive real numbers, prove that $$\frac{a}{\sqrt{(a + 2b)^3}}+\frac{b}{\sqrt{(b + 2c)^3}} +\frac{c} {\sqrt{(c + 2a)^3}} \ge \frac{1}{\sqrt{a + b + c}}$$Alexandru Mihalcu Since the inequality is homogenous, assume $a+b+c=1$. Note that $f(x)=x^{-3/2}$ is convex, so by weighted Jensen: $$\sum_{\text{cyc}}af(a+2b)\ge f(a^2+b^2+c^2+2ab+2bc+2ca)=f(1)=1.$$ sqing wrote: Let $a, b, c$ are positive real numbers. Prove that $$\frac{1}{\sqrt{a + 2b}}+\frac{1}{\sqrt{b + 2c}} +\frac{1} {\sqrt{c + 2a}} \ge \frac{3}{\sqrt{a + b + c}}$$ The same process with $f(x)=x^{-1/2}$ gives this one.

sqing wrote: Let $a, b, c$ are positive real numbers. Prove that $$\frac{1}{\sqrt{a + 2b}}+\frac{1}{\sqrt{b + 2c}} +\frac{1} {\sqrt{c + 2a}} \ge \frac{3}{\sqrt{a + b + c}}$$

Attachments:

$$\frac{a}{\sqrt{(a + 2b)^3}}+\frac{b}{\sqrt{(b + 2c)^3}} +\frac{c} {\sqrt{(c + 2a)^3}} \ge \frac{1}{\sqrt{a + b + c}}$$very good problem solved with $Yunis019$ $$\sum{\frac{a}{\sqrt{(a+2b)^3}}} \geq \frac{1}{\sqrt{a+b+c}} (?)$$$$\sum{\frac{a}{\sqrt{(a+2b)^3}}}+\sum{\frac{a}{\sqrt{a+2b}}} \geq \frac{1}{\sqrt{a+b+c}}+ \sum{\frac{a}{\sqrt{a+2b}}} (?)$$$$AM-GM$$$$\sum{\frac{a}{\sqrt{(a+2b)^3}}}+\sum{\frac{a}{\sqrt{a+2b}}} \geq 2\sum{\frac{a}{a+2b}} $$By homogenity we can say $a+b+c=1$ So $\sum{\frac{a}{a+2b}} \geq 1$ by titu's lemma only thing we need to prove is $$\sum{\frac{a}{a+2b}} \geq \sum{\frac{a}{\sqrt{a+2b}}} (?)$$By $GM-HM$ for $RHS$ $\sum{\frac{a}{\sqrt{a+2b}}} \le \sum{\frac{a^2+2ab+a}{2(a+2b)}}$ $$\sum{\frac{a}{a+2b}} \geq \sum{\frac{a^2+2ab+a}{2(a+2b)}} (?)$$$\implies$ $\sum{\frac{a}{a+2b}} \geq 1$ $\blacksquare$

sqing wrote: Let $a, b, c$ are positive real numbers. Prove that $$\frac{1}{\sqrt{a + 2b}}+\frac{1}{\sqrt{b + 2c}} +\frac{1} {\sqrt{c + 2a}} \ge \frac{3}{\sqrt{a + b + c}}$$ $$\frac{1}{\sqrt{a + 2b}}+\frac{1}{\sqrt{b + 2c}} +\frac{1} {\sqrt{c + 2a}} \ge \frac{3}{\sqrt{a + b + c}}$$By Titu $$\frac{1}{\sqrt{a + 2b}}+\frac{1}{\sqrt{b + 2c}} +\frac{1} {\sqrt{c + 2a}} \geq \frac{9}{\sum{\sqrt{a+2b}}}$$by cauchy for left $$\frac{9}{\sum{\sqrt{a+2b}}} \geq \frac{9}{\sqrt{9(a+b+c)}}=\frac{3}{\sqrt{(a+b+c)}}$$

By Hölder, \[\left(\sum_{\rm cyc}\frac a{\sqrt{(a+2b)^3}}\right)^2(a+b+c)\ge\left(\sum_{\rm cyc}\frac a{a+2b}\right)^3.\]Thus, it suffices to prove that $\frac{\left(\sum\limits_{\rm cyc}\frac a{a+2b}\right)^3}{a+b+c}\ge\left(\frac1{\sqrt{a+b+c}}\right)^2$, or $\sum_{\rm cyc}\frac a{a+2b}\ge1$. By C-S, \[\sum_{\rm cyc}\frac a{a+2b}=\sum_{\rm cyc}\frac{a^2}{a^2+2ab}\ge\frac{(a+b+c)^2}{\sum\limits_{\rm cyc}\left(a^2+2ab\right)}=1.\]We're done!