andreiromania wrote: Let $a,b,c,d \ge 0$ real numbers so that $a+b+c+d=1$.Prove that $\sqrt{a+\frac{(b-c)^2}{6}+\frac{(c-d)^2}{6}+\frac{(d-b)^2}{6}} +\sqrt{b}+\sqrt{c}+\sqrt{d} \le 2.$ By C-S $4=(1+3)\left(a+\frac{b-c)^2}{6}+\frac{(b-d)^2}{6}+\frac{(c-d)^2}{6}+b+c+d-\frac{(b-c)^2}{6}-\frac{(b-d)^2}{6}-\frac{(c-d)^2}{6}\right)\geq$ $\geq\left(\sqrt{(a+\frac{b-c)^2}{6}+\frac{(b-d)^2}{6}+\frac{(c-d)^2}{6}}+\sqrt{3(b+c+d)-b^2-c^2-d^2+bc+bd+cd}\right)^2$. Hence, it remains to prove that $3(b+c+d)-b^2-c^2-d^2+bc+bd+cd\geq\left(\sqrt b+\sqrt c+\sqrt d\right)^2$ or $2\sum_{cyc}(b-\sqrt{bc})\geq\sum_{cyc}(b^2-bc)$. Since $1\geq b+c+d$, it remains to prove that $2(b+c+d)\sum_{cyc}(b-\sqrt{bc})\geq\sum_{cyc}(b^2-bc)$, which is obvious.

China Girls Math Olympiad 2007 ,Problem 6 : For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that $ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$

For $a,b,c,d \ge 0$ with $a+b+c+d=1$.Prove that $$\sqrt{a+\frac{(b-c)^2}{6}+\frac{(c-d)^2}{6}+\frac{(d-b)^2}{6}} +\sqrt{b}+\sqrt{c}+\sqrt{d} \ge 1.$$

andreiromania wrote: Let $a,b,c,d \ge 0$ real numbers so that $a+b+c+d=1$.Prove that $\sqrt{a+\frac{(b-c)^2}{6}+\frac{(c-d)^2}{6}+\frac{(d-b)^2}{6}} +\sqrt{b}+\sqrt{c}+\sqrt{d} \le 2.$

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My solution to my problem, we have $$3(b+c+d)-\left(\sqrt b+\sqrt c+\sqrt d\right)^2=\left(\sqrt b-\sqrt c\right)^2+\left(\sqrt c-\sqrt d\right)^2+\left(\sqrt d-\sqrt b\right)^2.\qquad (1)$$But $$\frac{(b-c)^2}{2}=\frac{(\sqrt b+\sqrt c)^2(\sqrt b-\sqrt c)^2}{2}\le (b+c)(\sqrt b-\sqrt c)^2\le (\sqrt b-\sqrt c)^2$$and analogously $$\left(\sqrt c-\sqrt d\right)^2\ge \frac{(c-d)^2}{2}\;,\;\;\left(\sqrt d-\sqrt b\right)^2\ge \frac{(d-b)^2}{2}\,.$$Using them in $(1)$ we obtain $$\frac{\left(\sqrt b+\sqrt c+\sqrt d\right)^2}{3}\le b+c+d-\frac{(b-c)^2}{6}-\frac{(c-d)^2}{6}-\frac{(d-b)^2}{6}\qquad (2)$$ Applying the \emph{Cauchy-Schwarz} inequality we have $$\left(\sqrt{a+\frac{(b-c)^2}{6}+\frac{(c-d)^2}{6}+\frac{(d-b)^2}{6}}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)^2\le \left(a+\frac{(b-c)^2}{6}+\frac{(c-d)^2}{6}+\frac{(d-b)^2}{6}+\frac{\left(\sqrt b+\sqrt c+\sqrt d\right)^2}{3}\right)(1+3)\stackrel{(2)}{\le} 4(a+b+c+d)=4$$and such the inequality is proved. Equality holds when $a=b=c=d=\frac 14$.

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and a stronger one from Mathematical Reflections 4-2015: For $ a,b,c,d\geq 0$ with $ a+b+c+d=1$, prove that\[ \sqrt{a+\frac{2(b-c)^2}{9}+\frac{2(c-d)^2}{9}+\frac{2(d-b)^2}{9}}+\sqrt{b}+\sqrt{c}+\sqrt{d}\leq 2.\](Proposed by Marius Stanean, Zalau, Romania)

https://imomath.com/srb/zadaci/2019_opstinsko.pdf Generalization

MariusStanean wrote: and a stronger one from Mathematical Reflections 4-2015: For $ a,b,c,d\geq 0$ with $ a+b+c+d=1$, prove that\[ \sqrt{a+\frac{2(b-c)^2}{9}+\frac{2(c-d)^2}{9}+\frac{2(d-b)^2}{9}}+\sqrt{b}+\sqrt{c}+\sqrt{d}\leq 2.\] See my solution in this topic. It still works. The following inequality is also true. Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $a+b+c+d=1.$ Prove that: $$\sqrt{a+\frac{5}{11}(b^2+c^2+d^2-bc-bd-cd)}+\sqrt{b}+\sqrt{c}+\sqrt{d}\leq2.$$