Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 Rezolvarea problemelor din Revista de Matematica Mathematical Reflection – MR6/2013 J283 Solution by Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania We can write the inequality from the enounce in the equivalent form: 2a + 1 ∑ b+c = 2∑ a 1 9 +∑ ≥ 3+ 2∑ a b+c b+c From the Nesbitt inequality, we get : a 3 ∑b+c ≥ 2 1 ∑b+c ≥ 2 9 ∑a , By substitution we obtain the desired result. J285 Solution by Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania Using the known notation in a triangle, we have : a + b + c = 2 s, ab + bc + ca = s 2 + r 2 + 4 Rr , a 2 + b 2 + c 2 = 2s 2 − 2r 2 − 8Rr , abc = 4 Rrs Then the inequality from the enounce becomes : 2 s (2 s 2 + 2r 2 + 8Rr − 2s 2 + 2r 2 + 8Rr ≤9⇔ 4 Rrs 8 R < 2 r + 8 R ≤ 9 R ⇔ R ≥ 2r 8< which is the Euler inequality. Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 U283 Solution by Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania Consider the function f : R − > R, f ( x) = x n , which is convex. From the Tiberiu Popoviciu inequality, we have : a+b+c+d a+b+c+d −a a +b +c + d −b ) ≥ 3[ f ( )+ f ( )+ 4 3 3 a+b+c+d −c a+b+c+d −d )+ f ( )] ⇔ f( 3 3 1 2−a n 2−b n 2−c n 2−d n ) +( ) +( ) +( ) ]⇔ a n + b n + c n + d n + 8 n ≥ 3[( 2 3 3 3 3 2−a n 2−b n 2−c n 2−d n a n + b n + c n + d n + 2 3− n ≥( ) +( ) +( ) +( ) 3 3 3 3 3 f (a ) + f (b) + f (c) + f (d ) + 8 f ( ,q.e.d O283 Solution by Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania We denote S 21 = x22 + x32 + ... + xn2 , S 22 = x12 + x32 + ... + xn2 , ................................... S 2n = x12 + x22 + ... + xn2−1 , S 2 = x12 + x22 + ... + xn2 Suppose that : x1 ≥ x2 ≥ ..... ≥ xn , We shall prove that : Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 x13 x23 ≥ 2 ⇔ x13 ( x12 + x32 + ... + xn2 ) − x23 ( x22 + x32 + ... + xn2 ) ≥ 0 ⇔ 1 S2 S2 x15 − x25 + ( x13 − x23 ) x32 + ... + ( x13 − x23 ) xn2 ≥ 0 ⇔ ( x1 − x2 )( x14 + x13 x2 + .. + x24 ) + ... + ( x1 − x2 )( x12 + x1 x2 + x22 ) xn2 ≥ 0 ⇒ x1 ≥ x2 and S 21 ≤ S 22 ,similarly can be proven the others inequalities. It follows that the, n-uples ( xn3 x13 x23 , ,..., ), ( S 21 , S 22 ,..., S 2n ) ,are in the reversing order ,so that we can S 21 S 22 S 2n use the Cebisev: inequality 3 3 3 xn3 x13 x23 n xn 1 x1 2 x2 ( S + S + ... + S )( 1 + 2 + ... + n ) ≥ n( S2 1 + S2 2 + ... + S2 n ) ⇔ S 2 S2 S2 S2 S2 S2 1 2 n 2 2 2 (n − 1) S2 ( xn3 x13 x23 + + ... + ) ≥ n( x13 + x23 + ... + xn3 ) ⇔ S21 S22 S2n n x13 x23 xn3 n ∑ i =1 + + ... + ≥ S21 S22 S2n n − 1 n xi3 ∑x 2 i i =1 n n ∑ xi3 ⇔ i =1 n ∑x i i =1 ≥ n ∑x 2 i , (1) i =1 More generally, we shall prove the following inequality: Consider x1 , x2 ,..., xn > 0 and k > p ≥ 0 two natural numbers. Then : n n ∑x ∑x i =1 n i =1 k i i ≥( ∑x p i n )k − p , (*) i =1 Proof : We have the identity : Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 (n − 1)( x1k + x2k + ... + xnk ) = [( x1k −1 − x2k −1 )( x1 − x2 ) + ( x1k −1 − x3k −1 )( x1 − x3 ) + ... + ( xnk−−11 − xnk −1 )( xn −1 − xn )] + + ( x2k −1 + ... + xnk −1 ) x1 + ( x1k −1 + x3k −1 + .. + xnk −1 ) x2 + .. + ( x1k −1 + .. + xnk−−11 ) xn The expressions from the parantheses are positive, so that : ( n − 1)( x1k + x2k + ... + xnk ) ≥ ( x2k −1 + ... + xnk −1 ) x1 + ( x1k −1 + x3k −1 + .. + xnk −1 ) x2 + .. + ( x1k −1 + .. + xnk−−11 ) xn , whence ( n − 1)( x1k + x2k + ... + xnk ) ≥ ( x1k −1 + x2k −1 + ... + xnk −1 )( x1 + x2 + .. + xn ) − ( x1k + x2k + ... + xnk ) , si deci n(( x1k + x2k + ... + xnk ) ≥ ( x1k −1 + x2k −1 + ... + xnk −1 )( x1 + x2 + .. + xn ) We take for the exponents the values k-1,k-2,…,p+1 and we multiply the obtained inequalities. We get : n n n n k − p ∑ xik ≥ ( ∑ xip )(∑ xi ) k − p , whence it follows the inequality (*) i =1 i =1 i =1 In our case, for k=3 and p=2, we obtain the inequality (1). Q.e.d Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 Probleme propuse de matematica Nicusor Zlota O64 Solve the system in O65 Let Prove that such that . Solutie Let , , , , then we have to prove i.e. which is true by following weighted AM-GM O66 Let O67 such that . Prove that : Revista Virtuala Info MateTehnic Let ISSN 2069-7988 ISSN-L 2069-7988 .Prove that : i)If , then ii)If , then Generalizare O68 In .Prove that : O69 Let be positive reals. Prove the inequality Solutie By Titu's lemma we have Hence it suffices to prove Let then we have to prove which is obvious. , , , , Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 O70 Let ABC be an arbitrary triangle, and D, E, F any three points on the segments BC, CA, AB. Denote by P, Q, R the intersections of the lines AD, BE, CF with the circumcircle of triangle ABC (different from A, B, C). Show the inequality . When does equality hold? Solutie Denote by : 1. ; 2.suppose : ; 3.using Stewart's , we have : ; so, ; ; 4. We have : : ; By Q.E.D O71 In .Prove that : O72 In .Prove that : Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 Solutie After Ravi's substitution we need to prove that , which is obvious. O73 Solve the equation in R Discourse Solutie If and If and If no solution , the set of solutions is , and setting , equation is and so Which is a classical quartic which may be solved or discussed as a simple course question. O74 In , where .Prove that area of O75 Let Prove that : continous function such that Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 Solutie If , and for , we have: , and , conclusion is obvious O76 Let Solutie be non-negative real number,and only one can be zero. Prove that -1 Folosind metoda UVW, avem succesiv : Let , and . Hence, . Let For . We see that if then our inequality is true. it remains to prove that , which is true for Solutie-22 Utilizand metoda pqr si inegalitatea lui Schur Solutie-3 –Folosind inegalitatea cu numar minim de variabile, avem : suppose that . Let We have that . . Let , . Then we will prove that Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 It's equivalent to O77 Let be the angles of a triangle. Prove that Solutie : 1.we have : , 2. , 3.By AM-GM\Rightarrow Q.E.D O78 Let such that .Prove that Solution. Homogenizing we get Now by Cauchy-Schwarz, It suffice to prove or, equivalently, and simplifying turns this into AM-GM's: Expanding which is true by adding the following Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 O79 Let such that . Compute Solution : Well, you probably won't be given Note that substitution to unless it's relevant. Let and that . Hence, make the and add the two resulting integrals. The result is that the value is equal , which is trivial. O80 Let such that Find to maximum of Generalization : Solution Weighted AM-GM gives. But Hence Hence . . Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 Therefore O81 Prove that O82 Let be the sidelengths of triangle. Show that : O83 Let Prove that : and be sequence of complex. Nicusor Zlota-Focsani, Romania O84 Let such that .Prove that : Sol Notice so by AM-GM. Therefore and as desired. or , Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 Solution By AM GM we have . So \dfrac18 \ge abc a^2+b^2+c^2+2abc \ge 8abc 2abc$ completes the solution. The problem is equivalent to the: she returns to __________ Sandu Marin which is well-known! O85 Prove that the following inequality holds for all positive real numbers Solution Well, O86 Let Generalization . Prove that and : Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 Solution Chebyshev Cauchy-Schwarz so Cauchy-Schwarz and Muirhead. O87 Let Prove that : Solution It's equivalent to , which is Cauchy-Schwarz. O88 Let be the sidelengths of triangle. Show that : Revista Virtuala Info MateTehnic ISSN 2069-7988 ISSN-L 2069-7988 Solution : The left inequality is Muirhead. The right inequality is . Solution -2 by CS by CS by CS which is true again by CS. O89 Let such that . Prove that: O90 Let such that Solution AM-GM inequality helps here Indeed, we have We are done .Prove that :
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