Note that \begin{align*} (3a-1)^2 &\geqslant 0 \\ \Leftrightarrow (a+1)^2 &\geqslant 8a(1-a) \\ \Rightarrow (a+1)^2(2a(1-a)) &\geqslant 16a^2(1-a)^2 \\ \Rightarrow (a+1)\sqrt{2a(1-a)} &\geqslant 4a(1-a) = 4a(b+c). \end{align*}Thus, we have $$LHS \geqslant 4a(b+c)+4b(c+a)+4c(a+b) = RHS.$$

Equality at $(a,b,c)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$.

$$\frac{(2a)+(b+c)}{2}\geq\sqrt{2a(b+c)}$$$$\Rightarrow (2a+b+c)\sqrt{2a(b+c)}\geq 4a(b+c)$$Add the cyclic permutations to get the desired inequality. Equality at $2a=b+c$, $2b=a+c$ and $2c=a+b$, so $a=b=c$

Amir Hossein wrote: Positive real numbers $a,b,c$ satisfy $a+b+c=1$. Prove that $$(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \geq 8(ab+bc+ca).$$Also, find the values of $a,b,c$ for which the equality happens. See also here.

A Jensen's solution cause I am not that good at using AM-GM

Is this also gonna be compiled for the contest collections?

Amir Hossein wrote: Positive real numbers $a,b,c$ satisfy $a+b+c=1$. Prove that $$(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \geq 8(ab+bc+ca).$$Also, find the values of $a,b,c$ for which the equality happens. See also here https://artofproblemsolving.com/community/c6h1425240p8027908 Let $a,b,c $ be positive real numbers such that $ a+b+c=1.$ Prove thhat$$\frac{4\sqrt{2}}{3}\ge (1+a)\sqrt{a(1-a)}+(1+b)\sqrt{b(1-b)}+(1+c)\sqrt{c(1-c)} \ge 4\sqrt{2}(ab+bc+ca). $$

Devastator wrote: Is this also gonna be compiled for the contest collections? Greece JBMO TST 2017.

My solution: Notice $$2a+(1-a)=a+1 \ge 2\sqrt{2a(1-a)}$$by AM-GM. So the inequality is equivalent to $$(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \geq \sum_{cyc} 2(2a(1-a))=\sum_{cyc} 4a(1-a)= 8\sum_{cyc} ab.$$

sqing wrote: Amir Hossein wrote: Positive real numbers $a,b,c$ satisfy $a+b+c=1$. Prove that $$(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \geq 8(ab+bc+ca).$$Also, find the values of $a,b,c$ for which the equality happens. See also here. Sorry to ask but can you explain to me Why does the last equality work ? from $$\sqrt{2a(b+c)}\sqrt{2a(b+c)}$$to $$4(ab+ac)$$isn't it $$2(ab+ac)$$

The main claim is $(a+1)\sqrt{2a(1-a)}\ge4a(1-a)$, after which the problem is trivial. Indeed, it's true iff: $$2a(a+1)^2(1-a)\ge16a^2(1-a)^2,$$$$\Leftrightarrow(a+1)^2\ge8a(1-a),$$$$\Leftrightarrow 9a^-6a+1\ge0,$$$$\Leftrightarrow(3a-1)^2\ge0.$$ For the upper bound of $\text{LHS}\le\frac83$, we can use Jensen. The function $f(x)=(x+1)\sqrt{2x(1-x)}$ is concave since $f''(x)=\frac{8x^3-12x^2+3x-1}{2(x(1-x))^{3/2}\sqrt2}$ and $8x^3-12x^2+3x-1\le8x^2-12x^2+3x-1=-\left(x-\frac38\right)^2-\frac7{64}<0$. Then: $$f(a)+f(b)+f(c)\le3f\left(\frac{a+b+c}3\right)=\frac83.$$