How prove this $ \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}\geq 1.$ Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$, Prove that


  $$ \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}\geq 1.$$


This problem is from  http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&amp;t=555716

@Calvin Lin  Thank you
Let $\displaystyle A = \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}$ and $\displaystyle B = \sum_{cyc}a^2(a+3b+5bc)$.

Then by Hölder's inequality we have $A^2B \ge (a+b+c)^3 = 27$.

So it is sufficient to prove that $B \le 27$

$$B = \sum_{cyc}a^3 + 3 \sum_{cyc}a^2b+5\sum_{cyc}a^2bc$$ 

As $\displaystyle \sum_{cyc}ab^2 \ge 3abc$ by AM-GM, we have

$$B \le \left(\sum_{cyc}a^3 + 3 \sum_{cyc}a^2b + 3 \sum_{cyc}ab^2 + 6abc\right) - 15abc + 5\sum_{cyc}a^2bc \\
 = (a+b+c)^3 - 5abc (3-\sum_{cyc}a) = 27$$
We know the one result if sum of two numbers $a$ and $b$ is constant, the max. of $ab$ occurs when both $a$ and $b$ equal. Eg. if $a+b=6$, then $\max\{ab\}=9$. Use this result, we get proof.