JustPostTaiwanTST wrote: Let $ a,b,c,d $ be four non-negative reals such that $ a+b+c+d = 4 $. Prove that $$ a\sqrt{3a+b+c}+b\sqrt{3b+c+d}+c\sqrt{3c+d+a}+d\sqrt{3d+a+b} \ge 4\sqrt{5} $$ By Holder and C-S we obtain: $$\sum_{cyc}a\sqrt{3a+b+c}=\sqrt{\frac{\left(\sum\limits_{cyc}a\sqrt{3a+b+c}\right)^2\sum\limits_{cyc}\frac{a}{3a+b+c}}{\sum\limits_{cyc}\frac{a}{3a+b+c}}}\geq\sqrt{\frac{(a+b+c+d)^3}{\sum\limits_{cyc}\frac{a}{3a+b+c}}}=$$$$=\sqrt{\frac{64}{\frac{4}{3}+\sum\limits_{cyc}\left(\frac{a}{3a+b+c}-\frac{1}{3}\right)}}=\sqrt{\frac{64}{\frac{4}{3}-\frac{1}{3}\sum\limits_{cyc}\frac{b+c}{3a+b+c}}}\geq\sqrt{\frac{64}{\frac{4}{3}-\frac{1}{3}\frac{\left(\sum\limits_{cyc}(b+c)\right)^2}{\sum\limits_{cyc}(b+c)(3a+b+c)}}}=$$$$=\sqrt{\frac{48}{1-\frac{16}{\sum\limits_{cyc}(3ab+3ac+2a^2+2ab)}}}=\sqrt{\frac{48}{1-\frac{32}{\sum\limits_{cyc}(4a^2+10ab+6ac)}}}=$$$$=\sqrt{\frac{48}{1-\frac{32}{5(a+b+c+d)^2-(a-c)^2-(b-d)^2}}}\geq\sqrt{\frac{48}{1-\frac{32}{5(a+b+c+d)^2}}}=4\sqrt5.$$

Let $ a,b,c,d $ be four non-negative reals such that $ a+b+c+d = 4 $. Prove that$$ a\sqrt{3a+b+c}+b\sqrt{3b+c+d}+c\sqrt{3c+d+a}+d\sqrt{3d+a+b} \le 8\sqrt{3}. $$

sqing wrote: Let $ a,b,c,d $ be four non-negative reals such that $ a+b+c+d = 4 $. Prove that$$ a\sqrt{3a+b+c}+b\sqrt{3b+c+d}+c\sqrt{3c+d+a}+d\sqrt{3d+a+b} \le 8\sqrt{3}. $$ Solution. Using the Cauchy-Schwarz's Inequality and the condition $a+b+c+d=4$, we get \begin{align*}&\left(a\sqrt{3a+b+c}+b\sqrt{3b+c+d}+c\sqrt{3c+d+a}+d\sqrt{3d+a+b}\right)^2\\ =&\left(\sqrt a\sqrt{a(3a+b+c)}+\sqrt b\sqrt{b(3b+c+d)}+\sqrt c\sqrt{c(3c+d+a)}+\sqrt d\sqrt{d(3d+a+b)}\right)^2\\ \le&(a+b+c+d)\left[a(2a+4-d)+b(2b+4-a)+c(2c+4-b)+d(2d+4-c)\right]\\ \le&4\left[a(2a+4)+b(2b+4)+c(2c+4)+d(2d+4)\right]\\ =&4\left[2\left(a^2+b^2+c^2+d^2\right)+4(a+b+c+d)\right]\\ \le&4\left[2\left(a+b+c+d\right)^2+4\cdot4\right]=8^2\cdot3, \end{align*}which gives the desired inequality. $\blacksquare$

Let $ a,b,c $ be non-negative reals such that $ a+b+c = 3 .$ Prove that$$ a\sqrt{3a+b}+b\sqrt{3b+c}+c\sqrt{3c+a}\le 9 .$$Solution. Using the Cauchy-Schwarz's Inequality and the condition $a+b+c=3$, we get \begin{align*}&\left(a\sqrt{3a+b}+b\sqrt{3b+c}+c\sqrt{3c+a}\right)^2\\ =&\left(\sqrt a\sqrt{a(3a+b)}+\sqrt b\sqrt{b(3b+c)}+\sqrt c\sqrt{c(3c+a)}\right)^2\\ \le&(a+b+c)\left[a(3a+b)+b(3b+c)+c(3c+a)\right]\\ =&3\left[27- 5(ab+bc+ca)\right]\le 9^2 , \end{align*}which gives the desired inequality. $\blacksquare$ Let $ a,b,c $ be non-negative reals such that $ a+b+c = 3 .$ Prove that$$ 6\le a\sqrt{3a+b}+b\sqrt{3b+c}+c\sqrt{3c+a}\le 9 .$$

Let $ a_1,a_2,\cdots,a_n (n\ge 3) $ be non-negative reals such that $ a_1+a_2+\cdots+a_n = n .$ Prove that$$n\sqrt{n+1}\le a_1\sqrt{2a_1+S-a_n}+a_2\sqrt{2a_2+S-a_1}+a_3\sqrt{2a_3+S-a_2}+\cdots+a_n\sqrt{2a_n+S-a_{n-1}}\le n\sqrt{3n}. $$Where $S=a_1+a_2+\cdots+a_n.$