The following inequality is also true Let $a,b,c\in \mathbb{R}^+$, prove that: $$\frac{a}{\sqrt{2a+b}}+\frac{b}{\sqrt{2b+c}}+\frac{c}{\sqrt{2c+a}}\leq \sqrt{a+b+c}$$

Yes! C-S and C-S.

mihajlon wrote: Let $a,b,c\in \mathbb{R}^+$, prove that: $$\frac{2a}{\sqrt{3a+b}}+\frac{2b}{\sqrt{3b+c}}+\frac{2c}{\sqrt{3c+a}}\leq \sqrt{3(a+b+c)}$$ By Cauchy-Schwarz, we have that $RHS^2 \le \left( \displaystyle\sum_{cyc} \dfrac{a}{3a+b} \right) \left( 4a+4b+4c \right)$. It suffices to show this quantity is $\le 3(a+b+c)$, or equivalently that $\displaystyle\sum_{cyc} \dfrac{a}{3a+b} \le \dfrac{3}{4}$. But this is equivalent with $\displaystyle\sum_{cyc} \dfrac{b}{3a+b} \ge \dfrac{1}{4}$. By Cauchy-Schwarz again, $LHS = \displaystyle\sum_{cyc} \dfrac{b^2}{3ab+b^2} \ge \dfrac{ (a+b+c)^2}{a^2+b^2+c^2+3ab+3bc+3ca} \ge \dfrac{3}{4}$, where the final step comes from expanding and making use of $a^2+b^2+c^2 \ge ab+bc+ca$.

rmrf wrote: mihajlon wrote: Let $a,b,c\in \mathbb{R}^+$, prove that: $$\frac{2a}{\sqrt{3a+b}}+\frac{2b}{\sqrt{3b+c}}+\frac{2c}{\sqrt{3c+a}}\leq \sqrt{3(a+b+c)}$$ By Cauchy-Schwarz, we have that $RHS^2 \le \left( \displaystyle\sum_{cyc} \dfrac{a}{3a+b} \right) \left( 4a+4b+4c \right)$. It suffices to show this quantity is $\le 3(a+b+c)$, or equivalently that $\displaystyle\sum_{cyc} \dfrac{a}{3a+b} \le \dfrac{3}{4}$. But this is equivalent with $\displaystyle\sum_{cyc} \dfrac{b}{3a+b} \ge \dfrac{1}{4}$. By Cauchy-Schwarz again, $LHS = \displaystyle\sum_{cyc} \dfrac{b^2}{3ab+b^2} \ge \dfrac{ (a+b+c)^2}{a^2+b^2+c^2+3ab+3bc+3ca} \ge \dfrac{3}{4}$, where the final step comes from expanding and making use of $a^2+b^2+c^2 \ge ab+bc+ca$. Very nice. $\displaystyle\sum_{cyc} \dfrac{b}{3a+b}= \displaystyle\sum_{cyc} \dfrac{b^2}{3ab+b^2} \ge \dfrac{ (\sum_{cyc} a)^2}{\sum_{cyc} a^2+3\sum_{cyc} ab} \ge \dfrac{3}{4}\implies \displaystyle\sum_{cyc} \dfrac{a}{3a+b} \le \dfrac{3}{4}$, $\sum_{cyc} \frac{a}{\sqrt{3a+b}}\le \sqrt{\left( \displaystyle\sum_{cyc} \dfrac{a}{3a+b} \right) \left( a+b+c \right)}\leq \frac{1}{2}\sqrt{3(a+b+c)}$.

luofangxiang wrote: The following inequality is also true Let $a,b,c\in \mathbb{R}^+$, prove that: $$\frac{a}{\sqrt{2a+b}}+\frac{b}{\sqrt{2b+c}}+\frac{c}{\sqrt{2c+a}}\leq \sqrt{a+b+c}$$ Expanding the whole thing and using Cauchy Schwartz inequality is a key idea !!

mihajlon wrote: Let $a,b,c\in \mathbb{R}^+$, prove that: $$\frac{2a}{\sqrt{3a+b}}+\frac{2b}{\sqrt{3b+c}}+\frac{2c}{\sqrt{3c+a}}\leq \sqrt{3(a+b+c)}$$ Let $a,b\in \mathbb{R}^+.$ Prove that: $$\frac{a}{\sqrt{3a+b}}+\frac{b}{\sqrt{3b+a}}\leq \sqrt{\frac{a+b}{2}}\leq\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3a}}.$$Let $a,b,c\in \mathbb{R}^+ .$ Prove that: $$\frac{a}{\sqrt{3a+b}}+\frac{b}{\sqrt{3b+c}}+\frac{c}{\sqrt{3c+a}}\leq \frac{1}{2}\sqrt{3(a+b+c)}\leq\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3a}}.$$

Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers. Prove that $$\frac{a_1}{\sqrt{3a_1+a_2}}+\frac{a_2}{\sqrt{3a_2+a_3}}+\cdots+\frac{a_{n-1}}{\sqrt{3a_{n-1}+a_n}}+\frac{a_n}{\sqrt{3a_n+a_1}}\leq \frac{1}{2}\sqrt{n(a_1+a_2+\cdots+a_n)}\leq \frac{a_1}{\sqrt{a_1+3a_2}}+\frac{a_2}{\sqrt{a_2+3a_3}}+\cdots+\frac{a_{n-1}}{\sqrt{a_{n-1}+3a_n}}+\frac{a_n}{\sqrt{a_n+3a_1}}.$$

rmrf wrote: mihajlon wrote: $\displaystyle\sum_{cyc} \dfrac{a}{3a+b} \le \dfrac{3}{4}$. But this is equivalent with $\displaystyle\sum_{cyc} \dfrac{b}{3a+b} \ge \dfrac{1}{4}$. How do you know?

rmrf wrote: mihajlon wrote: $\displaystyle\sum_{cyc} \dfrac{a}{3a+b} \le \dfrac{3}{4}$. But this is equivalent with $\displaystyle\sum_{cyc} \dfrac{b}{3a+b} \ge \dfrac{1}{4}$. How do you know? Because $3\sum_{cyc}\frac{a}{3a+b} + \sum_{cyc}\frac{b}{3a+b} = 3$, it may be $\sum_{cyc}\frac{b}{3a+b}\geq \frac{3}{4}$