<s>[INST] Let $a,b,c$ be the nonnegative real numbers such that $a+b+c=1$. Prove that $\sqrt{a+\frac{(b-c)^2}4}+\sqrt b+\sqrt c\le\sqrt3$ 
Let $a,b,c$ be the nonnegative real numbers such that $a+b+c=1$. Prove that
  $$\sqrt{a+\frac{(b-c)^2}4}+\sqrt b+\sqrt c\le\sqrt3$$

I first wrote $a$ as $1-b-c$ and substituted it in main inequality
$$\sqrt{4(1-b-c)+(b-c)^2}+2\sqrt b+2\sqrt c\le2\sqrt 3$$
I tried to find some relation between this inequality and QAGH inequalities, but I couldn't. Then I come up with an idea to square main inequality several times to cancel square roots and then to reduce it to sum of squares, but I am sure there is a better way to prove it. What is the best way to prove it?
 [/INST] I use Cauchy-Schwarz inequality we have
$$\left(\sqrt{a+\dfrac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\right)^2
\le \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)(1+2)
$$
$$\Longleftrightarrow \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)\le 1$$
since $1-a=b+c$
$$\Longleftrightarrow \dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\le b+c$$
$$\Longleftrightarrow \dfrac{(b-c)^2}{4}\le\dfrac{(\sqrt{b}-\sqrt{c})^2}{2}$$
$$\Longleftrightarrow (\sqrt{b}+\sqrt{c})^2\le 2$$
It is clear,Because use Cauchy-schwarz inequality 
$$b+c\le 1\Longrightarrow (\sqrt{b}+\sqrt{c})^2\le[1+1](b+c)$$
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