Olympiad Inequality: Cauchy Schwartz Can someone please help me with some hint or the solution of this problem below:
For $a,b,c,k >0$
$ \frac{a}{\sqrt{ka+b}} +\frac{b}{\sqrt{kb+c}} + \frac{c}{\sqrt{kc+a}} < \sqrt{\frac{(k+1)}{k}(a+b+c)}$ 
I tried the following but got stuck. 
By Cauchy Schwartz:
$ \frac{a}{\sqrt{ka+b}} +\frac{b}{\sqrt{kb+c}} + \frac{c}{\sqrt{kc+a}} \leq \sqrt{\frac{(a+b+c)}{k}}(\Sigma_{cyc} \frac{ka}{ka+b})^{1/2} $
(note I have multiplied and divided by $k$ in the right hand side in order to the form of the answer)
Now I have to prove :
$\Sigma_{cyc} \frac{ka}{ka+b} < k+1$ to get to the answer. But I am unable to prove that. 
 A: Another way:
By C-S
$$\sum_{cyc}\frac{a}{\sqrt{ka+b}}=\frac{\sum\limits_{cyc}a\sqrt{(kb+c)(kc+a)}}{\prod\limits_{cyc}\sqrt{ka+b}}\leq$$
$$\leq\frac{\sqrt{\sum\limits_{cyc}a^2(kb+c)\sum\limits_{cyc}(kc+a)}}{\prod\limits_{cyc}\sqrt{ka+b}}=\frac{\sqrt{\sum\limits_{cyc}(ka^2b+a^2c)(k+1)(a+b+c)}}{\prod\limits_{cyc}\sqrt{ka+b}}.$$
Id est, it's enough to prove that
$$k\sum_{cyc}(ka^2b+a^2c)\leq(ka+b)(kb+c)(kc+a)$$ or
$$\sum_{cyc}(k^2a^2b+ka^2c)\leq(k^3+1)abc+\sum_{cyc}(k^2a^2b+ka^2c)$$ or
$$(k^3+1)abc\geq0.$$
A: The hint for my old solution:
For $k\geq1$ use the following C-S:
$$\left(\sum_{cyc}\frac{a}{\sqrt{ka+b}}\right)^2\leq(a+b+c)\sum_{cyc}\frac{a}{ka+b}$$ and it's enough to prove the inequality, which you got:
$$\sum_{cyc}\frac{ka}{ka+b}\leq k+1$$ or
$$(k+1)(ka+b)(kb+c)(kc+a)\geq\sum_{cyc}ka(kb+c)(kc+a)$$ or
$$(k+1)(k^3+1)abc+\sum_{cyc}((k^3+k^2)a^2b+(k^2+k)a^2c)\geq$$
$$\geq\sum_{cyc}(k^3abc+k^2a^2b+k^2a^2b+ka^2c)$$ or
$$(k^4-2k^3+k+1)abc+\sum_{cyc}((k^3-k^2)a^2b+k^2a^2c)\geq0,$$ which is obvious.
For $0<k\leq1$ use C-S again:
$$\left(\sum_{cyc}\frac{a}{\sqrt{ka+b}}\right)^2\leq\sum_{cyc}\frac{a}{(ka+b)(ka+c)}\sum_{cyc}a(ka+c).$$
