2023 WSMO Speed Round Problems/Problem 7

Problem

Let $e, a, j$ be real numbers such that $e + a + j = 1$ and $e\geq -\frac{1}{3}$ , $a\geq 1$ and $j\geq-\frac{5}{3}$ . Find the maximum value of $\sqrt{3e+1} + \sqrt{3a+3} + \sqrt{3j+5}.$

Solution

From the Cauchy-Schwarz inequality, we have $\begin{align*} \left(\sqrt{3e+1}+\sqrt{3a+3}+\sqrt{3j+5}\right)^2&\le\left((3e+1)+(3a+3)+(3j+5)\right)\left(1+1+1\right)\\ &\le\left(3(e+a+j)+9\right)(3) = (3(1)+9)(3)=36\implies\\ \sqrt{3e+1}+\sqrt{3a+3}+\sqrt{3j+5}\le\sqrt{36}=\boxed{6}. \end{align*}$

~pinkpig

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2023 WSMO Speed Round Problems/Problem 7

Problem

Solution