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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
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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 :
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(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
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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 :
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Let
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.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.
,
,
,
,
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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
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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
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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
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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
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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
.
.
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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
,
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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
:
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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 :
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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 :