answer:1. **Start with the given condition:** [ x^2 + y^2 + z^2 + w^2 = 1 ] 2. **Use the AM-GM inequality (Arithmetic Mean - Geometric Mean Inequality):** The AM-GM inequality states for any non-negative real numbers (a_1, a_2, ldots, a_n): [ frac{a_1 + a_2 + cdots + a_n}{n} geq sqrt[n]{a_1 a_2 cdots a_n} ] Applying this to (x, y, z, w): [ frac{x + y + z + w}{4} geq sqrt[4]{x y z w} ] 3. **Apply the QM-AM inequality (Quadratic Mean - Arithmetic Mean Inequality):** QM-AM inequality states for any real numbers (a_1, a_2, ldots, a_n): [ sqrt{frac{a_1^2 + a_2^2 + cdots + a_n^2}{n}} geq frac{a_1 + a_2 + cdots + a_n}{n} ] Applying it to (x, y, z, w): [ sqrt{frac{x^2 + y^2 + z^2 + w^2}{4}} geq frac{x + y + z + w}{4} ] 4. **Substitute the given condition (x^2 + y^2 + z^2 + w^2 = 1):** [ sqrt{frac{1}{4}} = frac{1}{2} ] Therefore, combining the inequalities obtained from AM-GM and QM-AM: [ sqrt[4]{x y z w} leq frac{x+y+z+w}{4} leq frac{1}{2} ] 5. **Square both sides to isolate (x y z w):** [ x y z w leq left(frac{1}{2}right)^4 = frac{1}{16} ] 6. **Next, sum the inequalities:** By also considering: [ x+y+z+w leq 2 ] 7. **Multiply ( x y z w leq frac{1}{16} ) by ( x + y + z + w leq 2 ):** [ x y z w cdot (x + y + z + w) leq frac{1}{16} cdot 2 = frac{1}{8} ] 8. **Express the original sum in terms of ( x y z w cdot (x + y + z + w) ):** [ x^2 y z w + x y^2 z w + x y z^2 w + x y z w^2 = x y z w cdot (x + y + z + w) ] 9. **Conclusion:** [ x^2 y z w + x y^2 z w + x y z^2 w + x y z w^2 leq frac{1}{8} ] This completes the proof, thus: (boxed{frac{1}{8}})

answer:1. **Substitute for ( y ) in the second equation:** [ x + y = 3 implies y = 3 - x. ] Substituting ( y = 3 - x ) into ( y = frac{9}{x^2 + 3} ), we get: [ 3 - x = frac{9}{x^2 + 3}. ] 2. **Multiply both sides by ( (x^2 + 3) ) to clear the fraction:** [ (3 - x)(x^2 + 3) = 9. ] Expanding the left side: [ 3x^2 + 9 - x^3 - 3x = 9. ] Rearranging terms gives: [ x^3 - 3x^2 + 3x = 0. ] 3. **Factor out common terms:** [ x(x^2 - 3x + 3) = 0. ] Solving the quadratic ( x^2 - 3x + 3 = 0 ) using the quadratic formula: [ x = frac{-(-3) pm sqrt{(-3)^2 - 4 cdot 1 cdot 3}}{2 cdot 1} = frac{3 pm sqrt{9 - 12}}{2} = frac{3 pm sqrt{-3}}{2}. ] Since (sqrt{-3}) indicates no real roots, the only real solution comes from ( x = 0 ). 4. **Verify the solution ( x = 0 ):** Substituting ( x = 0 ) into the original equations: [ y = frac{9}{0^2 + 3} = frac{9}{3} = 3, ] and [ x + y = 0 + 3 = 3. ] Both equations are satisfied. 5. **Conclusion:** The value of ( x ) at the intersection of the given curves is ( 0 ). boxed{The final answer is ( boxed{B} )}

answer:1. Let ABC be a triangle with D as the foot of the A-altitude. The circle w with diameter [AD] intersects AB at K and AC at L. 2. Denote M as the intersection of the tangents to the circle w at points K and L. 3. We need to prove that the ray [AM] is the A-median in triangle ABC. First, we establish that the quadrilateral AKRL is harmonic and that [AS] is the A-symmedian in triangle KAL. 4. Since AD is the diameter of the circle w, AD is perpendicular to BC at D. Therefore, AD is the altitude of triangle ABC. 5. The power of point A with respect to the circle w gives us: [ AK cdot AB = AD^2 = AL cdot AC ] This implies that AK cdot AB = AL cdot AC, meaning that the quadrilateral BKLC is cyclic. 6. Since BKLC is cyclic, we have: [ angle BKL = angle BCL quad text{and} quad angle KBL = angle KCL ] This implies that triangle AKL sim triangle ACB by AA similarity. 7. Now, consider the tangents from K and L to the circle w. Let S be the intersection of KL and AM, and R be the second intersection of AM with the circle w. 8. By the properties of harmonic division, the quadrilateral AKRL is harmonic. This means that the cross-ratio (A, S; R, M) is harmonic. 9. Since AKRL is harmonic, the ray [AS] is the A-symmedian in triangle KAL. This follows from the fact that the symmedian is the isogonal conjugate of the median in a triangle. 10. Given that triangle AKL sim triangle ACB and [AS] is the A-symmedian in triangle KAL, it follows that [AM] is the A-median in triangle ABC. Therefore, we have shown that the ray [AM] is indeed the A-median in triangle ABC. blacksquare

answer:Let's analyze each proposition: For the first proposition, when a=0, there is no corresponding pure imaginary number; therefore, it is incorrect because it fails to establish a one-to-one correspondence for all real numbers. For the second proposition, although the product of two conjugate complex numbers is a real number, the converse is not necessarily true. There are non-conjugate complex numbers whose product is a real number. Thus, this proposition is incorrect. For the third proposition, the statement does not specify whether x and y are real numbers. Therefore, we cannot conclude that x=y=1 is the only solution for x+yi=1+i, making the condition sufficient but not necessary. Hence, this proposition is correct. For the fourth proposition, we cannot compare the magnitude of a real number with an imaginary number in the sense of "greater than" or "less than", so this proposition is not correct. Therefore, the correct choice is: boxed{C} By considering the special case of a=0 for the first proposition, applying the definition of conjugate complex numbers and their operations for the second proposition, recognizing that x and y don't necessarily have to be real numbers for the third proposition, and understanding that imaginary numbers cannot be ordered for the fourth proposition, we can solve this problem. This question examines the basic concepts of complex numbers and mastering these concepts is key to solving the problem, categorizing it as a fundamental question.