If the maximum value of 5-x-\\frac{1}{2x} for x \\in \\mathbb{R}^+ can be expressed as a-\\sqrt{b}, where a, b \\in \\mathbb{Z}^+, find a+b.
", "hints_md": "Silver $=$ AG $=$ AM-GM", "hints_html": "Silver = AG = AM-GM
", "editorial_md": "For this question we will use the (unweighted) AM-GM inequality. The inequality states: $$\\frac{x_1 + x_2 + \\cdots + x_n}{n} \\geq \\sqrt[n]{x_1 x_2 \\cdots x_n} $$\n\nfor positive real numbers. Equality (when the two terms are equal) holds only when $x_1=x_2=x_3=\\cdots=x_n$. You can read further about this inequality [here](https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality), and you can read proofs [here](https://artofproblemsolving.com/wiki/index.php/Proofs_of_AM-GM).\n\nIn this question, we are finding the maximum value of $5-(a+\\frac{1}{2a})$, which is the same as finding the minimum value of $a+\\frac{1}{2a}$. Note that the equality holds when the two terms are equal, so we just need to solve $a=\\frac{1}{2a}$. You can then substitute this back into our original equation and find the answer.\n\n- fireheartjerry", "editorial_html": "For this question we will use the (unweighted) AM-GM inequality. The inequality states: \\frac{x_1 + x_2 + \\cdots + x_n}{n} \\geq \\sqrt[n]{x_1 x_2 \\cdots x_n}
for positive real numbers. Equality (when the two terms are equal) holds only when x_1=x_2=x_3=\\cdots=x_n. You can read further about this inequality here, and you can read proofs here.
In this question, we are finding the maximum value of 5-(a+\\frac{1}{2a}), which is the same as finding the minimum value of a+\\frac{1}{2a}. Note that the equality holds when the two terms are equal, so we just need to solve a=\\frac{1}{2a}. You can then substitute this back into our original equation and find the answer.