Revista Virtuala Info MateTehnic ISSN 69-7988 ISSN-L 69-7988 Probleme propuse spre rezolvare Solution by Mathematical Reflections and Mathematical Excalibu Nicusor Zlota O Let such that. Prove that O Solve the equation : Solution We need. Setting then, equation becomes Hence the two solutions O Solve the equation O Solve the equation for integer. O6 Let such that.prove that
O7 Let such that, find min Solution Mathematical Reflections MR/4 J We have : ab ( + ) = + ( + ) + a b a ab a b b using the assumption, we ab + + + ( + ) + ( + ) a b a ab a b b ab a b ab ( + ) a b J4. Analyzing the enounce we see that we have a problem of extremum with conditions defined by the function : f ( abc,, ) = a+ b+ c, and the link : F( abc,, ) = a+ b+ c = We shall find the extremum points of the function f(a,b,c) with the link F(a,b,c)=., considering the Lagrange function L( abc,,, λ) = f ( abc,, ) + λf( abc,, ), and calculating their stationary points. In these points the partial derivatives of the first order of the function L must be zero:
L = + λ = a L = + λ = b b L = + λ = c c a+ b+ c= This system has the solution : λ=, b=, c=, a= 4 4 α = (,,, ) 4 4, which is the stationary point of the function L. It follows that the function f has the stationary point with conditions : A (,, ) 4 4, corresponding to the Lagrange multiplier λ= The differential of second order of the Lagrange function in an arbitrary point of the domain of definition is : L L L L L L =, =, =, = = = a b 4b b c 9c c a b b c c a d L= da db dc () 4b b 9c c But, by differentiatind the link it follows :, da+db+dc=, we get dc=-db-da and replacing in (), we obtain : The hessian of this differential is : H ( ) L α, and has the principal minors <, >, so that, by the Sylvester criterion d L( α ) is a negatively definite quadratic form. The preceding results prove that A (,, ) 4 4 is a point of maximum with conditions of the function f; therefore we have : 5 f ( abc,, ) f (,, ) = + + = + = M 4 4 4 4 Similarly, we have
g abc a b c F abc a b c (,, ) = + +, (,, ) = + + = L ( abc,, ) = g( abc,, ) + λf( abc,, ) L = + λ= a L = + λ = b b+ c L = + λ= c 6 c b+ c a+ b+ c= λ=, a= +, b=, c= 4 4 β = ( +,,, ), B( +,, ) 4 4 4 4 5 g( abc,, ) g( +,, ) = + + = + = M 4 4 4 4 So, in conclusion we M= M S We denote by b = x a,a# c = y a, then the inequality becomes : Let x+ y= s, xy= p, then we ( + x+ y)( x+ xy+ y)( + x + y ) ( + x + y ) + s p + s s+ p + s ps ( ) ( )( )( ) Consider the function ) we show that : s f R f p = + s p + s s+ p + s ps 4 :[, ], ( ) ( ) ( )( )( ) f = + s + s s + s = s s + s + s s+ 6 5 4 () ( ) ( )( )( ) Calculate ' 5 4 f () = 6s 5s + 8s + 4s ; '' 4 f () = s s + 4s + 4 ''' f s s s s s s () = 6 + 48 = ( 5 + 4)
, true for any s ) Calculate : s s s s 6 5 4 f ( ) = ( + s ) ( + s)( s+ )(+ s s ) = ( s 5s + 8s 4s + 4s 6s+ 6) = 4 4 4 6 s s s +, 6 4 ( ) ( 4), true for any s S ) Let c= max{ abc,, }, we show that : x a+ x b > x c, x [,] a x b x Consider the function f :[,] R, f ( x) = ( ) + ( ), Is decreasing, so f ( x) > f (), c c a+ b x [,]. But f () =, we have f ( x ) >, for any x [,], result c x x x a+ b > c, x [,],. For x=, obtain a+ b > c ' ' ' ) Cosine theorem in the triangle ABC, we get ' b+ c a ' c+ a b ' a+ b c cos A =,cos B =,cosc = bc ac ab ' ' ' ( b+ c a)( c+ a b)( a+ b c) (p a)(p b)(p c) cosa cosb cosc = = = 8abc 8abc ( p a)( p b)( p c) ( p b)( p c) ( p a)( p c) ( p a)( p b) A = = sin, abc bc ac ab, where p=a+b+c
U Applying Jensen s inequality to the convex function f ( t) =, t>, we have : t x y z t x y z t + + + = f ( ) + f ( ) + f ( ) + f ( ) f ( + + + ) = x y z t 4 4 6 6 4 4 6 6 44 44 44 44 = = 6x+ 9y+ 4z+ t x+ 6y+ z+ x+ ( x+ y) + ( x+ y+ z) + ( x+ y+ z+ t) x+ 6y+ z+ 44 () + 6(4) + (6) + 44 U5 Let ( xn ) be a sequence n x = n... a + a + + a, which is strictly increasing, so ( x ) has n n n limited. If lim n xn = l, l R, then : lim n ( xn xn ) =,() But n n n n n = + +... + > > = an+ an+ an an+ + an+ +... + an a+ a+... + an 4n 4 x x, which contradicts (). Remains lim n x n =
Solution by MR4/4 J From the AM-GM inequality, we have ab bc c a abc a + c + a + b + (a + ) ( + ) ( + ) ( + ) ( ) + + We shall prove that : abc a ( ) a + + ( ) a a + bb + cc + a + b + c + ( ) ( ) ( ) ( )( )( ),(*) We have a a a a ( + ) + ( ) bb b b ( + ) + ( ) cc c c ( + ) + ( ) Multiplying,we obtain the inequality (*) O7 We see that : (,,, ),( a+, b+, c+, d+ ), are in the reverse order. a+ b+ c+ d+ From the Chebyshef inequality, we have: ( a+ + b+ + c+ + d+ )( + + + ) 4(( a+ ) + ( b+ ) + ( c+ ) + ( d+ ) ) a+ b+ c+ d+ a+ b+ c+ d+
( + a)( ) 6 a + a+ It suffices to prove that : abcd abcd By the means inequality it follows that a b c d abcd abcd 4 + + + 4, which is obvious. O In trianglel ABC we use the notations:bc=a,ac=b,ab=c BX CY AZ m=, n=, p=, where m, n, p (,) and from the Ceva s theorem, we deduce : BC CA AB m n p mnp ( m)( n)( p) m n p = =,(*) In triangle CXY, from the theorem of cosinuses, we get : = + cos CXY XY CX CY CXCY C XY = a ( m) + bn ab( mn ) cosc ab( mn ) ab( mn ) cosc = ab( mn ) ( cos C) Similarly, we have : = + AYZ YZ AY AZ AYAZ cosa YZ = b ( n) + c p bc( n) p cosa bc( n) p bc( n) p cosc = bc( n) p( cos A) = + BZX ZX BZ BX BZBX cosb ZX = c ( p) + a m ac( p) mcosa ac( p) m ac( p) mcosa = ac( p) m( cos A) Using (*) and the above relations, the inequality becomes
XBYC ZA R ( mnp) ( abc) R XY YZ ZX r 8( abc) ( mnp)( m)( n)( p)( cos A)( cos B)( cos C) 4r A B C 8( s a) ( s b) ( s c) r ( cos )( cos )( cos ) 8sin sin sin ( abc) R A B C = = = S7 Take D BC, then in triangle ADB, we have succesively : ADB AD= AB sinb a= 4csinB From the theorem of sinuses we get : sina= 4sinC sinb sin(4 B) = 4sin( B 6) sinb sin( B) cos( B) = 4sinB sin(6 B) ( cosb+ sin B)( sinb cos B) = 4sin B(sinB cos B) B B B B ( 4) sin + (+ 4 ) sin cos cos = Denoting tgb=t, the equation becomes : ( 4) t ( 4 ) t t + + = = +, whence tgb= + B= 75
S We shall prove that : b + c + d bcd, which is true. bcd b + c + d Similarly, we have c + d + a cda cda c + d + a d + a + b dab dab d + a + b a + b + c abc a + b + c abc The inequality is equivalent with : 6 b + c + d = bcd + a We see that : a + b + c + d + (,,, ),( a +, b +, c +, d + ), are in the reverse order. From the Chebyshef s inequality, we have : + + + + + + + + + + a + b + c + d + 4(( a + ) + ( b + ) + ( c + ) + ( d + ) ) a + b + c + d + 6 + a + a + 9 ( a b c d )( ) (8 a )( ) 6
Solution by Mathematical Excalibur /4 Problem 446. If real numbers a and b satisfy + = 7 and 5 + 7 =, then prove that <. a b a b If a b, then and 5 5. a b a a a a Then the relation : + = 7 + 7 Considering the function : a a f ( a ) = ( ) + ( ), Which is strictly decreasing and from 7 7 6 f () = <,we obtain f ( a) > > f (), therefore a<. 7 a b b b b b The relation 5 + 7 = 5 + 7 5 b 7 b Proceeding as above with the function : g( b ) = ( ) + ( ), we obtain b>, therefore b> > a which contradicts a b. It follows that a<b Generalization : x y x Let α, β, µ, σ (, ) and x, y R be real numbers such that α + β = ( α+ β+ ) and x y y µ + σ = ( µ + σ ).Show that x<y