answer:Given the conditions: [ a, b, c text{ are positive real numbers such that } a+b+c=1. ] We are tasked with proving that for all positive real numbers ( x, y, z ), the following inequality holds: [ left(x^{2}+y^{2}+z^{2}right)left(frac{a^{3}}{x^{2}+2 y^{2}}+frac{b^{3}}{y^{2}+2 z^{2}}+frac{c^{3}}{z^{2}+2 x^{2}}right) geq frac{1}{9}. ] 1. Start with the given inequality and rearrange it: [ left(x^{2}+y^{2}+z^{2}right)left(frac{a^{3}}{x^{2}+2 y^{2}}+frac{b^{3}}{y^{2}+2 z^{2}}+frac{c^{3}}{z^{2}+2 x^{2}}right) geq frac{a+b+c}{9} ] Since (a + b + c = 1), this simplifies to: [ left(x^{2}+y^{2}+z^{2}right)left(frac{a^{3}}{x^{2}+2 y^{2}}+frac{b^{3}}{y^{2}+2 z^{2}}+frac{c^{3}}{z^{2}+2 x^{2}}right) geq frac{1}{9} ] 2. Define ( P = frac{a^{3}}{x^{2}+2y^{2}} + frac{b^{3}}{y^{2}+2z^{2}} + frac{c^{3}}{z^{2}+2x^{2}} ) and ( N = 3 ( x^2 + y^2 + z^2 ) ). 3. Apply the AM-GM inequality (Arithmetical Mean-Geometrical Mean Inequality) to each fraction: [ frac{frac{a^{3}}{x^{2}+2y^{2}}}{P} + frac{x^{2}+2y^{2}}{N} + frac{1}{3} geq 3 sqrt[3]{frac{frac{a^{3}}{x^{2}+2y^{2}} cdot (x^{2}+2y^{2}) cdot frac{1}{3}}{P cdot N cdot 1}} ] Simplify each of these inequalities to: [ frac{a^{3}}{P (x^{2}+2y^{2})} + frac{x^{2}+2y^{2}}{3 (x^{2}+y^{2}+z^{2})} + frac{1}{3} geq 1 ] 4. Apply the above step similarly for ( b ) and ( c ): [ frac{frac{b^{3}}{y^{2}+2z^{2}}}{P} + frac{y^{2}+2z^{2}}{N} + frac{1}{3} geq frac{3b}{sqrt[3]{3 P N}} ] [ frac{frac{c^{3}}{z^{2}+2x^{2}}}{P} + frac{z^{2}+2x^{2}}{N} + frac{1}{3} geq frac{3c}{sqrt[3]{3 P N}} ] 5. Adding these inequalities: [ frac{frac{a^{3}}{x^{2}+2y^{2}}}{P} + frac{frac{b^{3}}{y^{2}+2z^{2}}}{P} + frac{frac{c^{3}}{z^{2}+2x^{2}}}{P} + frac{x^{2}+2y^{2}}{N} + frac{y^{2}+2z^{2}}{N} + frac{z^{2}+2x^{2}}{N} + 1 geq frac{3(a+b+c)}{sqrt[3]{3 P N}} ] 6. Simplifying both sides: [ 1 + 1 + 1 geq frac{3}{sqrt[3]{3 P N}} ] 7. Thus, we can get: [ boxed{P geq frac{(a+b+c)^3}{3N} = frac{1}{3 times 3 (x^2 + y^2 + z^2)}} ] Since ( P = frac{a^3}{x^2 + 2y^2} + frac{b^3}{y^2 + 2z^2} + frac{c^3}{z^2 + 2x^2} ): [ P geq frac{1}{9 (x^2 + y^2 + z^2)} ] Therefore, the following inequality holds true: [ left( x^2 + y^2 + z^2 right) left( frac{a^3}{x^2 + 2y^2} + frac{b^3}{y^2 + 2z^2} + frac{c^3}{z^2 + 2x^2} right) geq frac{1}{9} ] Conclusion: [ boxed{} ]

answer:Xia started with 150 stickers. After sharing 100 stickers with her friends, she had: 150 stickers - 100 stickers = 50 stickers left. Since each sheet contains 10 stickers, to find out how many sheets Xia had left, we divide the remaining stickers by the number of stickers per sheet: 50 stickers ÷ 10 stickers/sheet = 5 sheets. Xia had boxed{5} sheets of stickers left.

answer:Let (O) be the circumcenter of the tetrahedron (ABCD). Define the midpoints of the sides (BC), (CA), (AB) as (L), (M), and (N) respectively. We are given the equalities: 1. ( AB + BC = AD + CD ) 2. ( CB + CA = BD + AD ) 3. ( CA + AB = CD + BD ) Let's analyze these conditions more closely by adding the first two and subtracting the third: [ (AB + BC) + (CB + CA) - (CA + AB) = (AD + CD) + (BD + AD) - (CD + BD) ] Simplifying this, we get: [ BC + CB = AD + AD ] [ 2BC = 2AD ] So: [ BC = AD ] Similarly, adding the second and third and subtracting the first: [ (CB + CA) + (CA + AB) - (AB + BC) = (BD + AD) + (CD + BD) - (AD + CD) ] Simplifying this, we have: [ 2CA = 2BD ] Thus: [ CA = BD ] Finally, adding the first and third and subtracting the second: [ (AB + BC) + (CA + AB) - (CB + CA) = (AD + CD) + (CD + BD) - (BD + AD) ] Thus: [ 2AB = 2CD ] So: [ AB = CD ] With the equalities (AB = CD), (BC = AD), and (CA = BD) established, consider points on the opposite edges of tetrahedron (ABCD). Let ((L', M', N')) be the midpoints of ((AD, BD, CD)) respectively: [ L' = text{midpoint of } AD, quad M' = text{midpoint of } BD, quad N' = text{midpoint of } CD ] It can be shown that: 1. (L'M' = frac{1}{2} AB = LM) 2. (L'M') and (LM) are parallel to (AB), hence parallel and coplanar. Also: [ L'M = frac{1}{2}CD = frac{1}{2}AB = LM ] Thus, quadrilateral (LML'M') forms a rhombus. Let (X) be the intersection of the diagonals of this rhombus. Diagonal intersections in a rhombus indicate orthogonal bisectors, meaning (X) is: 1. Right angle intersection point of (LL') and (MM') 2. Midpoint of each diagonal and thus perpendicular to the plane of rhombus. Since (NN') is similarly a bisector perpendicular to the same plane and contains the circumcenter (O), (X) must lie at (O). Due to the perpendicularity: 1. (angle LOM = 90^circ) 2. (angle MUN = 90^circ) 3. (angle NOL = 90^circ) Thus, we conclude: [ angle LOM = angle MON = angle NOL = 90^circ ] (boxed{angle LOM = angle MON = angle NOL = 90^circ})

answer:1. **Identify Ratios from Given Data**: Given: [ frac{AK}{BK} = frac{3}{2} quad text{and} quad frac{BM}{MC} = frac{3}{1} ] This implies: [ BK = frac{2}{5} AB quad text{and} quad BM = frac{3}{4} BC ] 2. **Construct a Parallel Line**: Through point ( B ), draw a line ( l ) parallel to ( AC ). 3. **Intersecting Points and Similar Triangles**: The line ( KM ) intersects ( l ) at point ( P ) and ( AC ) at point ( N ). By similarity of triangles: [ triangle CMN sim triangle BMP ] Since ( l parallel AC ), the angles at ( C ) and ( B ) are corresponding angles, and their other pairs of angles at ( M ) are vertically opposite and equal. 4. **Ratios from Similar Triangles**: From the similarity of ( triangle CMN ) and ( triangle BMP ): [ frac{BP}{CN} = frac{BM}{MC} = 3 ] Hence: [ BP = 3 cdot CN ] 5. **Second Pair of Similar Triangles**: Considering triangles ( triangle AKN ) and ( triangle BKP ): [ triangle AKN sim triangle BKP ] From their corresponding sides: [ frac{AN}{KN} = frac{AK}{BK} = frac{3}{2} ] 6. **Applying Ratios**: Now, solving for ( AN ) using the given similarity and ( BP ) relation: [ AN = frac{3}{2} BP ] Since ( BP = 3 cdot CN ): [ AN = frac{3}{2} times 3 cdot CN = frac{9}{2} CN ] 7. **Express ( AC ) in Terms of ( CN )**: Given ( AC = a ): [ AC = AN + NC = frac{9}{2} cdot CN + CN = frac{11}{2} CN ] Thus: [ a = frac{11}{2} CN implies CN = frac{2a}{11} ] Subsequently: [ BP = 3 cdot CN = 3 cdot frac{2a}{11} = frac{6a}{11} ] Conclusion: [ boxed{frac{6a}{11}, frac{2a}{11}} ]