answer:Proof: Since x > y, we have x - y > 0; Because 2x+ frac {1}{x^{2}-2xy+y^{2}} - 2y = 2(x-y) + frac {1}{(x-y)^{2}} = (x-y) + (x-y) + frac {1}{(x-y)^{2}}; Furthermore, (x-y) + (x-y) + frac {1}{(x-y)^{2}} geq 3sqrt[3]{(x-y)^{2} cdot frac {1}{(x-y)^{2}}} = 3, equality holds when x-y=1; Therefore, 2x+ frac {1}{x^{2}-2xy+y^{2}} - 2y geq 3, which means 2x+ frac {1}{x^{2}-2xy+y^{2}} geq 2y+3. Thus, we have proved that 2x+ frac {1}{x^{2}-2xy+y^{2}} geq 2y+3, which can be encapsulated as boxed{2x+ frac {1}{x^{2}-2xy+y^{2}} geq 2y+3}.

answer:For the quadratic equation 16x^2 + ax + b = 0 to have exactly one solution, its discriminant must be zero: [ Delta = a^2 - 4 cdot 16 cdot b = 0. ] This simplifies to: [ a^2 - 64b = 0. ] Given the additional condition a^2 = 4b, substitute this into the discriminant condition: [ 4b - 64b = 0, ] [ -60b = 0. ] Therefore, ( b = 0 ). Substituting ( b = 0 ) back into ( a^2 = 4b ) gives: [ a^2 = 0, ] [ a = 0. ] Thus, the solutions are ( a = 0 ) and ( b = 0 ). Conclusion: The only solution is ( a = 0 ) and ( b = 0 ), leading to the equation ( 16x^2 = 0 ), which indeed has exactly one solution, ( x = 0 ). Thus, ( boxed{(a, b) = (0, 0)} ).

answer:Let's denote the number of adults as A. We know that the meal can cater to either 70 adults or 90 children. This means that the amount of food required for one adult is equivalent to the amount of food required for 90/70 = 9/7 children. If 21 adults have their meal, the amount of food consumed by them would be equivalent to the amount of food for 21 * (9/7) children. 21 * (9/7) = 3 * 9 = 27 children. So, the food consumed by 21 adults is equivalent to the food for 27 children. Now, we know that after these 21 adults have eaten, there is enough food left for approximately 63 children. The total food was initially enough for 90 children. After feeding the equivalent of 27 children (21 adults), we have: 90 - 27 = 63 children worth of food left. Since this matches the number of children that can be catered with the remaining food, it means that no other adults have eaten. Therefore, the number of adults that went for trekking is boxed{21} .

answer:1. **Isolate the square root term**: Define u = sqrt{x + 16}. The equation then transforms to: [ u - frac{8}{u} = 4 ] 2. **Eliminate the fraction by multiplying by u** (assuming u neq 0): [ u^2 - 8 = 4u ] Rearranging: [ u^2 - 4u - 8 = 0 ] 3. **Factorize or solve the quadratic equation**: Using the quadratic formula, u = frac{4 pm sqrt{16 + 32}}{2} = frac{4 pm sqrt{48}}{2} = frac{4 pm 4sqrt{3}}{2} = 2 pm 2sqrt{3}. 4. **Convert back to x**: - For u = 2 + 2sqrt{3}: [ sqrt{x + 16} = 2 + 2sqrt{3}, quad x + 16 = (2 + 2sqrt{3})^2 = 4 + 8sqrt{3} + 12, quad x = 20 + 8sqrt{3} ] - For u = 2 - 2sqrt{3} (this is valid since 2sqrt{3} approx 3.46 < 4): [ sqrt{x + 16} = 2 - 2sqrt{3}, quad x + 16 = (2 - 2sqrt{3})^2 = 4 - 8sqrt{3} + 12, quad x = 20 - 8sqrt{3} ] 5. **Conclusion**: We find x = 20 + 8sqrt{3} and x = 20 - 8sqrt{3}. Both roots can be checked and are valid, solving the adjusted equation without any extraneous roots: [ x = 20 + 8sqrt{3 text{ and } x = 20 - 8sqrt{3}} ] The final answer is C) boxed{x = 20 + 8sqrt{3} text{ and } x = 20 - 8sqrt{3}}