answer:We are required to prove the inequality: [ a^{4}+b^{4}+c^{4}+ frac{a^{2}}{(b+c)^{2}}+ frac{b^{2}}{(c+a)^{2}}+ frac{c^{2}}{(a+b)^{2}} geqslant a b+b c+c a , quad text{for positive real numbers } a, b, text{ and } c. ] 1. Let us denote ( M = a^{4} + b^{4} + c^{4} ). Using the arithmetic mean-geometric mean inequality (AM-GM inequality), we have: [ frac{a^{4}}{M} + frac{1}{3} geqslant frac{2a^{2}}{sqrt{3M}} ] [ frac{b^{4}}{M} + frac{1}{3} geqslant frac{2b^{2}}{sqrt{3M}} ] [ frac{c^{4}}{M} + frac{1}{3} geqslant frac{2c^{2}}{sqrt{3M}} ] 2. Adding these inequalities, we obtain: [ left(frac{a^{4}}{M} + frac{1}{3}right) + left(frac{b^{4}}{M} + frac{1}{3}right) + left(frac{c^{4}}{M} + frac{1}{3}right) geqslant frac{2a^{2}}{sqrt{3M}} + frac{2b^{2}}{sqrt{3M}} + frac{2c^{2}}{sqrt{3M}} ] [ frac{M}{M} + 1 geqslant frac{2(a^{2} + b^{2} + c^{2})}{sqrt{3M}} ] 3. Simplifying, we get: [ 1 + 1 geqslant frac{2(a^{2}+b^{2}+c^{2})}{sqrt{3M}} ] [ 2 geqslant frac{2(a^{2}+b^{2}+c^{2})}{sqrt{3M}} ] [ sqrt{3M} geqslant a^{2} + b^{2} + c^{2} ] 4. Squaring both sides, we obtain: [ 3M geqslant (a^{2} + b^{2} + c^{2})^{2} ] [ M geqslant frac{1}{3} (a^{2} + b^{2} + c^{2})^{2} ] 5. Using the fact that ((a^2 + b^2 + c^2) geqslant (ab + bc + ca)), we get: [ M geqslant frac{1}{3} (ab + bc + ca)^{2} ] 6. Let ( N = frac{a^{2}}{(b+c)^{2}} + frac{b^{2}}{(c+a)^{2}} + frac{c^{2}}{(a+b)^{2}} ). [ P = 2(a+b+c), quad R = 2(ab + bc + ca) ] 7. Applying AM-GM inequality on each fraction, we get: [ frac{a^{2}}{N(b+c)^{2}} + frac{a(b+c)}{R} + frac{b+c}{P} geqslant frac{3a}{sqrt[3]{NRP}} ] [ frac{b^{2}}{N(c+a)^{2}} + frac{b(c+a)}{R} + frac{c+a}{P} geqslant frac{3b}{sqrt[3]{NRP}} ] [ frac{c^{2}}{N(a+b)^{2}} + frac{c(a+b)}{R} + frac{a+b}{P} geqslant frac{3c}{sqrt[3]{NRP}} ] 8. Adding these inequalities, we obtain: [ frac{N}{N} + frac{R}{R} + frac{P}{P} geqslant frac{3(a+b+c)}{sqrt[3]{NRP}} ] [ 1 + 1 + 1 geqslant frac{3(a+b+c)}{sqrt[3]{NRP}} ] [ 3 geqslant frac{3(a+b+c)}{sqrt[3]{NRP}} ] [ sqrt[3]{NRP} geqslant a + b + c ] 9. Simplifying, we get: [ NRP geqslant (a + b + c)^{3} ] [ N geqslant frac{(a+b+c)^{3}}{RP} = frac{(a+b+c)^{3}}{4(a+b+c)(ab+bc+ca)} = frac{(a+b+c)^{2}}{4(ab+bc+ca)} ] 10. Because ((a+b+c)^{2} geqslant 3(ab+bc+ca)), we have [ N geqslant frac{3(ab+bc+ca)}{4(ab+bc+ca)} = frac{3}{4} ] 11. Summing up (M) and (N), we have: [ M + N geqslant frac{1}{3} (ab+bc+ca)^{2} + frac{3}{4} ] 12. Using the AM-GM inequality, we obtain: [ sqrt{frac{1}{3} (ab + bc + ca)^{2} cdot frac{3}{4}} = ab + bc + ca ] # Conclusion: [ a^{4} + b^{4} + c^{4} + frac{a^{2}}{(b+c)^{2}} + frac{b^{2}}{(c+a)^{2}} + frac{c^{2}}{(a+b)^{2}} geq ab + bc + ca ] (boxed{}).

answer:# Part (a): Determine the area of overlap 1. **Identify the variables and relationships:** - Let ( m ) be the length of the segments ( BQ = AK = AR = BP ). - Let ( n ) be the length of the segments ( AL = BL = BN = AN ). - Given that ( m + n = 5 ). 2. **Apply the Law of Cosines:** - Consider the triangle ( triangle LAN ): [ LN^2 = AN^2 + AL^2 - 2 cdot AN cdot AL cdot cos(angle LAN) ] - Since ( angle LAN = 180^circ - angle LAK ), we have: [ cos(180^circ - angle LAK) = -cos(angle LAK) ] - Substituting ( cos(angle LAK) = frac{m}{n} ): [ LN^2 = 2n^2 - 2n^2 cdot left(-frac{m}{n}right) = 2n^2 + 2mn ] 3. **Solve for ( n ):** - Given ( LN = 5 ): [ 5^2 = 2n^2 + 2mn ] [ 25 = 2n^2 + 2mn ] - Using ( m + n = 5 ): [ n = frac{-m + sqrt{m^2 + 52}}{2} ] - Substitute ( n ) back into the equation: [ frac{-m + sqrt{m^2 + 52}}{2} + m = 5 ] [ m^2 + 52 = m^2 - 20m + 100 ] [ 20m = 48 implies m = 2.4 ] 4. **Calculate the area of overlap:** - The overlap area is the sum of the areas of triangles ( triangle QLB ) and ( triangle RAN ): [ text{Area} = frac{1}{2} cdot BQ cdot AL + frac{1}{2} cdot AR cdot BN = m = 2.4 ] # Part (b): Determine the length of the segment between the other 2 points of intersection, ( A ) and ( B ) 1. **Calculate ( n ):** - Using ( n = sqrt{m^2 + 1} ): [ n = sqrt{2.4^2 + 1} = sqrt{5.76 + 1} = sqrt{6.76} approx 2.6 ] 2. **Apply the Law of Cosines to find ( AB ):** - Consider the triangle ( triangle ALB ): [ AB^2 = AL^2 + BQ^2 - 2 cdot AL cdot BQ cdot cos(angle ALB) ] - Since ( angle ALB = angle LAK ): [ AB^2 = 2n^2 - 2n^2 cdot cos(angle LAK) ] [ AB^2 = 2n^2 - 2n^2 cdot frac{m}{n} ] [ AB^2 = 2n^2 - 2mn ] - Substitute ( n ) and ( m ): [ AB^2 = 2 cdot 2.6^2 - 2 cdot 2.6 cdot 2.4 ] [ AB^2 = 2 cdot 6.76 - 2 cdot 6.24 ] [ AB^2 = 13.52 - 12.48 = 1.04 ] [ AB = sqrt{1.04} = frac{sqrt{26}}{5} ] The final answer is ( boxed{frac{sqrt{26}}{5}} )

answer:**Analysis** By transforming 2na_n=(n-1)a_{n-1}+(n+1)a_{n+1} for n geqslant 2 and n in mathbb{N}^*, we can deduce that na_n-(n-1)a_{n-1}=(n+1)a_{n+1}-na_n for n geqslant 2 and n in mathbb{N}^*. From this, we know that the sequence {na_n} is an arithmetic sequence with the first term being 1 and the common difference being 5. Thus, we can find that frac{a_n}{n} = frac{5n-4}{n^2} = frac{5- frac{4}{n}}{n} =(5- frac{4}{n} )cdot frac{1}{n}. By analyzing the monotonicity of the function f(x)=(5-4x)x, we can reach the conclusion. This problem tests the ability to find the general term of a sequence and the ability to solve it through calculation. Constructing a new sequence is the key to solving this problem. Pay attention to accumulating solving methods, as it is a medium-level question. **Solution** Since 2na_n=(n-1)a_{n-1}+(n+1)a_{n+1} for n geqslant 2 and n in mathbb{N}^*, It follows that na_n-(n-1)a_{n-1}=(n+1)a_{n+1}-na_n for n geqslant 2 and n in mathbb{N}^*, Given a_1=1, a_2=3, We have (n+1)a_{n+1}-na_n=na_n-(n-1)a_{n-1}=…=2a_2-a_1=5, Thus, the sequence {na_n} is an arithmetic sequence with the first term being 1 and the common difference being 5, Therefore, na_n=1+5(n-1)=5n-4, Hence, frac{a_n}{n} = frac{5n-4}{n^2} = frac{5- frac{4}{n}}{n} =(5- frac{4}{n} )cdot frac{1}{n}, Let f(x)=(5-4x)x, then the graph of y=f(x) is a downward-opening parabola symmetric about x= frac{5}{8}, Since 0 < x leqslant 1, the maximum value of f(x) is fleft( frac{5}{8} right)= frac{25}{16}, Given n in mathbb{N}^*, the maximum value of frac{a_n}{n} is achieved when n=2, which is frac{3}{2}, Therefore, the correct choice is boxed{text{B}}.

answer:Solution: (1) According to the problem, let A(x_{1},y_{1}) and B(x_{2},y_{2}), then we have y_{1}^{2}=8x_{1} and y_{2}^{2}=8x_{2}. Subtracting the two equations, we get (y_{1}-y_{2})(y_{1}+y_{2})=8(x_{1}-x_{2}). Since y_{1}+y_{2}=8, then k= dfrac {y_{2}-y_{1}}{x_{2}-x_{1}}=1, the slope of line AB is boxed{1}. (2) Given that F(2,0), the equation of line AB is y=x-2, substituting into y^{2}=8x to eliminate y and rearranging, we get x^{2}-12x+4=0, thus x_{1}+x_{2}=12, using the chord length formula, we find |AB|=(x_{1}+x_{2})+p=boxed{16}.