Prove that $ \sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} } \leq \frac{3}{2} $ For all $a, b, c>0$ and $a+b+c=3$ Prove that  $$ \sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} } \leq \frac{3}{2} $$
I tried cauchy-schwarz inequality for the L. H. S like and I get 
$ [\left( \sqrt{a} \right) ^{2}+\left( \sqrt{b} \right) ^{2}+\left( \sqrt{c} \right) ^{2}]\left[ \left( \frac{1}{\sqrt{b+3} } \right) ^{2}+\left( \frac{1}{\sqrt{c+3} } \right) ^{2}+\left( \frac{1}{\sqrt{a+3} } \right) ^{2}\right] \geq \left( \frac{\sqrt{a} }{\sqrt{b+3} } +\frac{\sqrt{b} }{\sqrt{c+3} } +\frac{\sqrt{c} }{\sqrt{a+3} } \right) ^{2}$
Then I get by AM-GM the maximum value of $abc=1$ and the inequality have the value 3... How I can prove inequality?. 
 A: First, we use Cauchy-Schwartz inequality:
$$
A^2=\left(\sqrt{\frac{a}{b+3} } +\sqrt{\frac{b}{c+3} } +\sqrt{\frac{c}{a+3} }\right)^2 =\\
\left(\sqrt{\frac{a}{(a+3)(b+3)}(a+3) } +\sqrt{\frac{b}{(b+3)(c+3)} (b+3)} +\sqrt{\frac{c}{(c+3)(a+3)}(c+3) }\right)^2 \leq \\
\left({\frac{a}{(a+3)(b+3)} } +{\frac{b}{(b+3)(c+3)} } +{\frac{c}{(c+3)(a+3)} }\right)\times (a+3+b+3+c+3) .
$$
Since $a+3+b+3+c+3=12$, it is enough to prove:
$$
\left({\frac{a}{(a+3)(b+3)} } +{\frac{b}{(b+3)(c+3)} } +{\frac{c}{(c+3)(a+3)} }\right)\leq \frac 3{16}. (\star)
$$
From this step on, I went the ugly way, since the manipulation was not so frightening. With a simple and nice proof of this step, the whole proof would become much easier.
Anyway, after the simple multiplications, we get to the following inequality:
$$
18-7(ab+ac+bc)+3abc\geq 0.
$$
This can be further simplified, using the fact such as $a^2+b^2+c^2=9-2(ab+ac+bc)$ and using the famous following identity:
$$
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).
$$ 
We can further simplify the inequality to the following:
$$
a^3+b^3+c^3\geq a^2+b^2+c^2.
$$
This follows from the following steps:


*

*Cauchy-Schwartz: $(a+b+c)(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2 \implies 3(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2$

*Cauchy-Schwartz: $(1+1+1)(a^2+b^2+c^2)\geq (a+b+c)^2 \implies (a^2+b^2+c^2)\geq 3.$

*Combine two previous steps:
$$
3(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2\geq  3(a^2+b^2+c^2).
$$
And therefore, this proves $(\star)$.

