28th Hellenic Mathematical Olympiad

We use the inequality of arithmetic–geometric mean as follows:

$$\sqrt[3]{a^2+2bc}=\frac{1}{\sqrt[3]{12^2}}\sqrt[3]{(a^2+2bc)\cdot 12\cdot 12}\le \frac{1}{\sqrt[3]{12^2}}\cdot \frac{a^2+2bc+12+12}{3}=\frac{1}{3\sqrt[3]{12^2}}(a^2+2bc+24),$$ $$\sqrt[3]{b^2+2ca}=\frac{1}{\sqrt[3]{12^2}}\sqrt[3]{(b^2+2ca)\cdot 12\cdot 12}\le \frac{1}{\sqrt[3]{12^2}}\cdot \frac{b^2+2ca+12+12}{3}=\frac{1}{3\sqrt[3]{12^2}}(b^2+2ca+24),$$ $$\sqrt[3]{c^2+2ab}=\frac{1}{\sqrt[3]{12^2}}\sqrt[3]{(c^2+2ab)\cdot 12\cdot 12}\le \frac{1}{\sqrt[3]{12^2}}\cdot \frac{c^2+2ab+12+12}{3}=\frac{1}{3\sqrt[3]{12^2}}(c^2+2ab+24),$$ from which we get $$\begin{array}{rl}S& =\sqrt[3]{a^2+2bc}+\sqrt[3]{b^2+2ca}+\sqrt[3]{c^2+2ab}\\ & \le \frac{1}{3\sqrt[3]{12^2}}(a^2+b^2+c^2+2ab+2bc+2ca+72)\\ & =\frac{1}{3\sqrt[3]{12^2}}\left[(a+b+c{)}^2+72\right]=\frac{36}{\sqrt[3]{12^2}}=\frac{18}{\sqrt[3]{18}}=3\sqrt[3]{12}.\end{array}$$ Equality holds when $$a^2+2bc=12,b^2+2ca=12,c^2+2ab=12$$ $$\iff (a-b)(a+b-2c)=0,(b-c)(b+c-2a)=0,c^2+2ab=12$$ $$\iff (a-b)(6-3c)=0,(b-c)(6-3a)=0,c^2+2ab=12$$ $$\iff a=b=c=2.$$ Therefore the maximal value of the expression is $3\sqrt[3]{12}$ and it is obtained for $a=b=c=2$.

Solution 2: Solution with Hölder's inequality (P. Lolas, G. Vlachos, M. Agelis)

We have $$\begin{array}{rl}\sqrt[3]{9(a+b+c{)}^2}& =\sqrt[3]{9\cdot \sqrt[3]{(a^2+2bc)+(b^2+2ca)+(c^2+2ab)}}\\ & =\sqrt[3]{1+1+1}\cdot \sqrt[3]{(a^2+2bc)+(b^2+2ca)+(c^2+2ab)}\\ & \ge \sqrt[3]{a^2+2bc}+\sqrt[3]{b^2+2ca}+\sqrt[3]{c^2+2ab}=S.\end{array}$$ Hence we have $S\le \sqrt[3]{9\cdot 6^2}=3\sqrt[3]{12}$, where equality holds for $$a^2+2bc=b^2+2ca=c^2+2ab,a+b+c=6$$ $$\iff (a-b)(a+b-2c)=0,(b-c)(b+c-2a)=0,a+b+c=6$$ $$\iff (a-b)(6-3c)=0,(b-c)(6-3a)=0,a+b+c=6$$ $$\iff a=b=c=2.$$

Solution 3: Solution using Cauchy–Schwarz inequality (X. Tsampasidis)

We put $x=\sqrt[3]{a^2+2bc}$, $y=\sqrt[3]{b^2+2ca}$, $z=\sqrt[3]{c^2+2ab}$. Then $x,y,z>0$ and from Cauchy–Schwarz inequality we get $$\begin{array}{rl}& (1^3+1^3+1^3)(1^3+1^3+1^3)(x^3+y^3+z^3)\ge (1\cdot 1\cdot x+1\cdot 1\cdot y+1\cdot 1\cdot z{)}^3\\ & \iff 9(x^3+y^3+z^3)\ge (x+y+z{)}^3.\end{array}$$ However we have $$x^3+y^3+z^3=a^2+2bc+b^2+2ca+c^2+2ab=(a+b+c{)}^2=36.$$ Thus the last inequality becomes $$x+y+z\le \sqrt[3]{9(x^3+y^3+z^3)}=\sqrt[3]{9\cdot 36}=3\sqrt[3]{12}.$$ Equality holds when $x=y=z,a+b+c=6$ or equivalently $$a^2+2bc=b^2+2ca=c^2+2ab,a+b+c=6\iff a=b=c=2.$$

Solution 4: Solution using Jensen's inequality (K. Axiotis, N Athanasiou, D. Sotiriou)

The function $f(x)=\sqrt[3]{x},x>0$, is twice differentiable in $(0,+\infty )$, with $f^{\prime }(x)=\frac{1}{3}{x}^{-\frac{2}{3}}$ and $f^{\prime \prime }(x)=-\frac{2}{9}{x}^{-\frac{5}{3}}$. Hence $f(x)=\sqrt[3]{x}$ is concave in $(0,+\infty )$. Hence from Jensen's inequality we get $$\begin{array}{rl}& \frac{1}{3}\left[f(a^2+2bc)+f(b^2+2ca)+f(c^2+2ab)\right]\le f\left(\frac{a^2+2bc+b^2+2ca+c^2+2ab}{3}\right)\\ \Rightarrow & f(a^2+2bc)+f(b^2+2ca)+f(c^2+2ab)\le 3f\left(\frac{(a+b+c{)}^2}{3}\right)=3f(12)=3\sqrt[3]{12}.\\ \Rightarrow & \sqrt[3]{a^2+2bc}+\sqrt[3]{b^2+2ca}+\sqrt[3]{c^2+2ab}\le 3\sqrt[3]{12}.\end{array}$$ Since for $a=b=c=2$ the first part takes the value $3\sqrt[3]{12}$, it follows that the maximal value of the expression $S$ is $3\sqrt[3]{12}$.

Solution 5: Solution using power means inequality (A. Mousatov)

We put $x=\sqrt{a^2+2bc}$, $y=\sqrt{b^2+2ca}$, $z=\sqrt{c^2+2ab}$. Then we have $$x+y+z=(a+b+c{)}^2=36.$$ Hence, if we put $u=\sqrt[3]{x}$, $v=\sqrt[3]{y}$, $w=\sqrt[3]{z}$, then we get $S=u+v+w$ and from power means inequality we obtain $$S=u+v+w\le 3\cdot \sqrt[3]{\frac{u^3+v^3+w^3}{3}}=3\cdot \sqrt[3]{12},$$ since $u^3+v^3+w^3=x+y+z=36$. Equality is valid when $$u=v=w\iff a=b=c=2.$$

Solution 6: Solution using rearrangement inequality (K. Dermentzis)

We put $x=\sqrt[3]{a^2+2bc}$, $y=\sqrt[3]{b^2+2ca}$, $z=\sqrt[3]{c^2+2ab}$. Then we have $$S=x+y+z\text{ and }{x}^3+y^3+z^3=(a+b+c{)}^2=36.$$ Next we put $$S_1=x^3+y^3+z^3=x^{2}x+y^{2}y+z^{2}z,S_2=x^{2}y+y^{2}z+z^{2}x,S_3=x^{2}z+y^{2}x+z^{2}y$$ And since the triads $(x^2,y^2,z^2)$, $(x,y,z)$ have the same ordering, from rearrangement inequality we obtain $S_1\ge S_2$ and $S_1\ge S_3$. Moreover, since $x,y,z>0$, we have $S_1=x^3+y^3+z^3\ge 3xyz$, and therefore $$\begin{array}{rl}{S}^3& =x^3+y^3+z^3+3(x^{2}y+y^{2}z+z^{2}x)+3(x^{2}z+y^{2}x+z^{2}y)+6xyz\\ & =S_1+3S_2+3S_3+6xyz\le S_1+3S_1+3S_1+2S_1=9S_1.\end{array}$$ Hence $$S=x+y+z\le \sqrt[3]{9(x^3+y^3+z^3)}=\sqrt[3]{9\cdot 12}=3\cdot \sqrt[3]{12}.$$ Since for $a=b=c=2$ the expression $S$ takes the value $3\sqrt[3]{12}$, it follows that its maximal value is $3\sqrt[3]{12}$.