Given $a+b+c=1$, prove that $\sqrt{a+\frac{(b-c)^2}4}+\sqrt{b}+\sqrt{c}\le \sqrt{3}$.
So far, I have tried to apply cauchy schwarz somehow because this works well with square roots and the inequality signs match up. However, this nonhomogeneity is tripping me up, so I would like to know how I could solve this inequality. Thanks!
By Cauchy-Schwarz inequality $$\left(\sqrt{a+\dfrac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\right)^2 \le \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)(1+2) $$ $$\Longleftrightarrow \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)\le 1$$ since $1-a=b+c$ $$\Longleftrightarrow \dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\le b+c$$ $$\Longleftrightarrow \dfrac{(b-c)^2}{4}\le\dfrac{(\sqrt{b}-\sqrt{c})^2}{2}$$ $$\Longleftrightarrow (\sqrt{b}+\sqrt{c})^2\le 2$$ and $$b+c\le 1\Longrightarrow (\sqrt{b}+\sqrt{c})^2\le[1+1](b+c)$$
