อสมการ
 

1). ให้ $ a,b,c>0 $ และ $a^2+b^2+c^2=1$ จงแสดงว่า
$ (\frac{1}{a^3(b+c)^5}+\frac{1}{b^3(c+a)^5}+\frac{1}{c^3(a+b)^5})^{\frac{1}{5}}\geqslant \frac{3}{2} $

2). $x,y,z>0$ จงแสดงว่า
$ \sqrt{\frac{x}{x+y}} + \sqrt{\frac{y}{y+z}} + \sqrt{\frac{z}{z+x}} \leqslant \frac{3}{\sqrt{2}} $

For 1) According to Holder inequality, we obtain
$(\sum_{cyc}\frac{1}{a^3(b+c)^5})(\sum_{cyc}a^2)^4\ge(\sum_{cyc}\frac{a}{b+c})^5$

For 2) solution by Mikhail:p. Applying Cauchy-Schwarz inequality
$\sum_{cyc}\sqrt{\frac{x}{x+y}}\le\sqrt{(\sum_{cyc}x+z)(\sum_{cyc}\frac{x}{(x+y)(x+z)})}$
It suffices to show that
$\sum_{cyc}\frac{x}{(x+y)(x+z)}\le\frac{9}{4(x+y+z)}$
You can expand to prove the last inequality.:)