China Girls' Mathematical Olympiad

Proof I Without loss of generality, we may assume that $b\ge c$. We set $\sqrt{b}=x+y$ and $\sqrt{c}=x-y$ for some nonnegative real numbers $x$ and $y$. Hence $b-c=4xy$ and $a=1-2x^2-2y^2$. It follows that $$\sqrt{a+\frac{(b-c{)}^2}{4}}+\sqrt{b}+\sqrt{c}=\sqrt{1-2x^2-2y^2+4x^2{y}^2}+2x.$$

Note that $2x=\sqrt{b}+\sqrt{c}$, implying that $$4x^2=(\sqrt{b}+\sqrt{c}{)}^2\le 2b+2c\le 2,$$ by the AM-GM inequality. Thus, $4x^2{y}^2\le 2y^2$ and $$1-2x^2-2y^2+4x^2{y}^2\le 1-2x^2.$$ Substituting the last inequality yields $$\begin{array}{rl}\sqrt{a+\frac{(b-c{)}^2}{4}}+\sqrt{b}+\sqrt{c}& \le \sqrt{1-2x^2}+2x\\ & =\sqrt{1-2x^2}+x+x\\ & \le \sqrt{3},\end{array}$$ by the Cauchy-Schwarz inequality.

Proof II Let $a=u^2$, $b=v^2$, and $c=w^2$. Then $u^2+v^2+w^2=1$ and the desired inequality becomes $$\sqrt{u^2+\frac{(v^2-w^2{)}^2}{4}}+v+w\le \sqrt{3}.$$

Note that $$\begin{array}{rl}{u}^2+\frac{(v^2-w^2{)}^2}{4}& =1-(v^2+w^2)+\frac{(v^2-w^2{)}^2}{4}\\ & =\frac{4-4(v^2+w^2)+(v^2-w^2{)}^2}{4}\\ & =\frac{4-4(v^2+w^2)+(v^2+w^2{)}^2-4v^2{w}^2}{4}\\ & =\frac{(2-v^2-w^2{)}^2-4v^2{w}^2}{4}\\ & =\frac{(2-v^2-w^2-2vw)(2-v^2-w^2+2vw)}{4}\\ & =\frac{[2-(v+w{)}^2][2-(v-w{)}^2]}{4}\\ & \le 1-\frac{(v+w{)}^2}{2}.\end{array}$$ (Note that $(v+w{)}^2\le 2(v^2+w^2)\le 2$)

Substitute the above equation and it gives $$\sqrt{1-\frac{(v+w{)}^2}{2}}+v+w\le \sqrt{3}.$$ Set $\frac{v+w}{2}=x$. We can rewrite the above inequality as $$\sqrt{1-2x^2}+2x\le \sqrt{3},$$ and we can complete the proof as we did in the first proof.