IMO Team Selection Contest

Using the AM-GM inequality for three terms twice, one gets $$\begin{array}{rl}\frac{2c}{a}+\frac{c}{b}& =\frac{(2b+a)c}{ab}=\frac{(2b+a)\sqrt{2a^2+b^2}}{3ab}=\frac{(b+b+a)\sqrt{a^2+a^2+b^2}}{3ab}\\ & \ge \frac{3\sqrt[3]{b^{2}a}\sqrt{3\sqrt[3]{a^4{b}^2}}}{3ab}=\frac{3\sqrt{3}\sqrt[3]{b^{2}a\cdot a^{2}b}}{3ab}=\frac{3\sqrt{3}ab}{3ab}=\sqrt{3}.\end{array}$$

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Alternative solution.

Using HM-QM inequality for $a,a,b$ gives $$\frac{3}{\frac{2}{a}+\frac{1}{b}}=\frac{3}{\frac{1}{a}+\frac{1}{a}+\frac{1}{b}}\le \sqrt{\frac{a^2+a^2+b^2}{3}}=\sqrt{\frac{2a^2+b^2}{3}}=\sqrt{\frac{9c^2}{3}}=\sqrt{3}c.$$ Thus $$\left(\frac{2}{a}+\frac{1}{b}\right)\cdot c\ge \frac{3}{\sqrt{3}}=\sqrt{3},$$ which implies the desired inequality.

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Alternative solution.

$$\begin{array}{rl}{\left(\frac{2c}{a}+\frac{c}{b}\right)}^2& =\frac{(2b+a{)}^2{c}^2}{a^2{b}^2}=\frac{(2b+a{)}^2(2a^2+b^2)}{9a^2{b}^2}=\\ & =\frac{((a^2+b^2)+4ab+3b^2)(a^2+(a^2+b^2))}{9a^2{b}^2}\ge \frac{(6ab+3b^2)(a^2+2ab)}{9a^2{b}^2}=\\ & =\frac{(2a+b)(a+2b)}{3ab}=\frac{2a^2+5ab+2b^2}{3ab}\ge \frac{4ab+5ab}{3ab}=3.\end{array}$$

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Alternative solution.

The square of the l.h.s. of the desired inequality is $${\left(\frac{2c}{a}+\frac{c}{b}\right)}^2=\frac{(2b+a{)}^2(2a^2+b^2)}{9a^2{b}^2}=\frac{1}{9}{\left(2+\frac{a}{b}\right)}^2\left(2+\frac{b^2}{a^2}\right).$$ Denoting $\frac{a}{b}=x$, the desired inequality reduces to $\frac{1}{9}(2+x{)}^2(2+\frac{1}{x^2})\ge 3$, which is equivalent to $(2+x{)}^2(2x^2+1)\ge 27x^2$. This in turn is equivalent to $2x^4+8x^3-18x^2+4x+4\ge 0$, that is $2(x-1{)}^2(x^2+6x+2)\ge 0$ after factorization. This inequality holds, since on positive arguments the quadratic polynomial $x^2+6x+2$ is positive.