whiwho wrote: Let $a,b,c$ be positive real numbers so that $a+b+c = \frac{1}{\sqrt{3}}$. Find the max of $27abc+a\sqrt{a^2+2bc}+b\sqrt{b^2+2ca}+c\sqrt{c^2+2ab}$ Solution of Zhangyanzong:

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Any other solutions please?

By Cauchy-Schwarz and AM-GM $$(27abc+a\sqrt{a^2+2bc}+b\sqrt{b^2+2ca}+c\sqrt{c^2+2ab})^2\le $$$$(9\sqrt{3}abc+a^2+b^2+c^2)(27\sqrt{3}abc+a^2+2bc+b^2+2ca+c^2+2ab)=$$$$\frac{1}{3} (27\sqrt{3}abc+3a^2+3b^2+3c^2)(27\sqrt{3}abc+a^2+b^2+c^2+2ab+2bc+2ca)\le$$$$\frac{1}{12}\left[54\sqrt{3}abc+4(a^2+b^2+c^2)+2(ab+bc+ca)\right]^2\le$$$$\frac{1}{12}\left[6(ab+bc+ca)+4(a^2+b^2+c^2)+2(ab+bc+ca)\right]^2=\frac{4}{27}.$$

Here's a Jensen solution First of all rewrite the conditions as $\sqrt{3}\cdot a+\sqrt{3}\cdot b+\sqrt{3}\cdot c=1$ and let the maximum value be $27abc+\sum_{cyc}a\sqrt{a^2+2b}\le A$ Which is also equivalent to $27abc\cdot\sqrt{3}+\sum_{cyc}\sqrt{3}\cdot a\sqrt{a^2+2bc}\le\sqrt{3}\cdot A$ Now let $x_1=a^2+2bc, x_2=b^2+2ca\text{ and }x_3=c^2+2ab$ thus $\sum_{cyc}\sqrt{3}\cdot a\sqrt{a^2+2bc}=\sum_{cyc}\sqrt{3}\cdot a\sqrt{x_1}$ Let $f(x)=\sqrt{x}=x^{1/2}$ therefore taking the first and second derivative yields $f'(x)=\frac{1}{2}x^{-1/2}\text{ and }f''(x)=-\frac{1}{4}x^{-3/2}<0, \forall x\in\mathbb{R}^+$ therefore $f$ is convex. Furthermore $27abc\cdot\sqrt{3}+\sum_{cyc}\sqrt{3}\cdot a\sqrt{a^2+2bc}=27abc\cdot\sqrt{3}+\sum_{cyc}\sqrt{3}\cdot a f(x_1)\overset{\text{W-Jensen}}{\le}27abc\cdot\sqrt{3}+f\left(\sqrt{3}\sum_{cyc}ax_1\right)=27abc\cdot\sqrt{3}+\sqrt{\sqrt{3}\left(\sum_{cyc}a^3+6abc\right)}$ Now notice that $27abc\overset{\text{AM-GM}}{\le}\left(\sum_{cyc}a\right)^3=\frac{1}{\sqrt[3]{3}^3}\text{ therefore }27abc\cdot\sqrt{3}\le\frac{\sqrt{3}}{\sqrt{3}^3}=\frac{1}{3}$ (1) Moreover $\sum_{cyc}a^3+6abc=\left(\sum_{cyc}a\right)^3-3\sum_{cyc}a\sum_{cyc}ab+9abc\le\frac{1}{\sqrt{3}^3}-\frac{3}{\sqrt{3}}\sum_{cyc}ab+\frac{1}{3\sqrt{3}^3}$ Therefore $\sqrt{3}\cdot\left(\sum_{cyc}a^3+6abc\right)\le\frac{\sqrt{3}}{\sqrt{3}^3}-3\sum_{cyc}ab+\frac{\sqrt{3}}{3\sqrt{3}^3}=\frac{4}{9}-3\sum_{cyc}ab$ Now we want to furthermore find the minimum of the last inequality and thus we need to find the minimum of $\sum_{cyc}ab$ so let's use Lagrange multipliers Let $f(a,b,c)=\sum_{cyc}ab, g(a,b,c)=\sum_{cyc}a-\frac{1}{\sqrt{3}}=0$ and $L=f(a,b,c)-\lambda g(a,b,c)$ $$\frac{\partial L}{\partial a}=b+c-\lambda=0\Longrightarrow b+c=\lambda$$$$\frac{\partial L}{\partial b}=a+c-\lambda=0\Longrightarrow a+c=\lambda$$$$\frac{\partial L}{\partial c}=b+a-\lambda=0\Longrightarrow b+a=\lambda$$Therefore $a+b=b+c=c+a=\lambda$ which forces $a=b=c$. Now plugging this in $g$ yields $g(a,a,a)=3a-\frac{1}{\sqrt{3}}=0\Longrightarrow a=\frac{1}{3\sqrt{3}}$ thus $a=b=c=\frac{1}{3\sqrt{3}}$ So $f\left(\frac{1}{3\sqrt{3}},\frac{1}{3\sqrt{3}},\frac{1}{3\sqrt{3}}\right)=3\left(\frac{1}{3\sqrt{3}}\right)^2=\frac{1}{9}$ which implies $\sum_{cyc}ab\ge\frac{1}{9}$ Now using this yields $\sqrt{3}\cdot\left(\sum_{cyc}a^3+6abc\right)\le\frac{\sqrt{3}}{\sqrt{3}^3}-3\sum_{cyc}ab+\frac{\sqrt{3}}{3\sqrt{3}^3}=\frac{4}{9}-3\sum_{cyc}ab\le\frac{4}{9}-\frac{3}{9}=\frac{1}{9}$ Therefore $\sqrt{\sqrt{3}\cdot\left(\sum_{cyc}a^3+6abc\right)}\le\sqrt{\frac{1}{9}}=\frac{1}{3}$ (2) Now summing the inequalities (1) and (2) yields $27abc\cdot\sqrt{3}+\sqrt{\sqrt{3}\left(\sum_{cyc}a^3+6abc\right)}\le\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$ So recall that $\frac{2}{3}\le\sqrt{3}A\Longrightarrow A\ge\frac{2}{3\sqrt{3}}$ thus the maximum is equal to $\boxed{\frac{2}{3\sqrt{3}}}$ $\blacksquare$.