mateforum.ro

In memoria domnului profesor Alexandru Lupas.

Inegalitatea lui Holder: Daca \( a_{1},a_{2}, \ldots,
a_{n} \), \( b_{1},b_{2}, \ldots, b_{n} \) si \( p,q \) numere reale pozitive astfel incat \( \frac{1}{p}+\frac{1}{q} = 1 \), atunci: \( \left(
\sum_{i=1}^{} a_{i}^{p} \right)^{\frac{1}{p}} \left( \sum_{i=1}^{}
b_{i}^{q} \right)^{\frac{1}{q}} \geq \sum_{i = 1}^{n} a_{i}b_{i}. \)

Aplicatii:

1. Daca \( a_{1},a_{2}, \ldots, a_{n} \); \( b_{1},
b_{2}, \ldots, b_{n} \), \( c_{1},c_{2}, \ldots, c_{n} \) si \( p,q,r \) sunt numere reale pozitive, cu \( \frac{1}{p}+\frac{1}{q}+\frac{1}{r} =1 \), atunci \( \left( \sum_{i=1}^{} a_{i}^{p} \right)^{\frac{1}{p}}
\left( \sum_{i=1}^{} b_{i}^{q} \right)^{\frac{1}{q}} \left(
\sum_{i=1}^{} c_{i}^{r} \right)^{\frac{1}{r}} \geq \sum_{i=1}^{n}
a_{i}b_{i}c_{i}. \)

(indicatie: mai intai se demonstreaza ca \( \frac{x^{p}}{p}+\frac{y^{q}}{q}+\frac{z^{r}}{r} \geq xyz \)).

Prezentam si alte exemple:

2. [Kiran Kedlaya]. Fie \( a,b,c \) numere reale pozitive. Sa
se arate ca \( \frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3} \leq
\sqrt[3]{a\left(\frac{a+b}{2}\right)\left(\frac{a+b+c}{3}
\right)}. \)

Solutie: Rezulta din primul exercitiu ca \( \left(
\sum_{i=1}^{3} a_{i}^3\right)^{\frac{1}{3}} \left( \sum_{i=1}^{3}
b_{i}^3 \right)^{\frac{1}{3}} \left( \sum_{i=1}^{3} c_{i}^3
\right)^{\frac{1}{3}} \geq \sum_{i=1}^{3} a_{i}b_{i}c_{i}(2) \)

Efectuand substitutiile \( a_{1} = a^{\frac{1}{3}}, a_{2} = a^{\frac{1}{3}}, a_{3} = a^{\frac{1}{3}}, b_{1} = a^{\frac{1}{3}}, b_{2} = \left( \frac{a+b}{2} \right)^{\frac{1}{3}}, b_{3} = b^{\frac{1}{3}}, c_{1} = a^{\frac{1}{3}}, c_{2} = b^{\frac{1}{3}}, c_{3} = c^{\frac{1}{3}} \) obtinem \( \sqrt[3]{a\left(\frac{a+b}{2}\right)\left(\frac{a+b+c}{3} \right)}
\geq \frac{a+\sqrt[3]{\frac{ab(a+b)}{2}}+\sqrt[3]{abc}}{3} \)

Cum \( \sqrt[3]{\frac{ab(a+b)}{2}} \geq \sqrt{ab} \), din inegalitatea mediilor, rezulta inegalitatea dorita.

3. [T. Andreescu, USAMO 2004] Fie \( a,b,c \) numere reale pozitive. Aratati ca: \( (a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3) \geq (a+b+c)^{3}. \)

Solutie Din inegalitatea lui Holder avem:
\( (a^3+1+1)^{\frac{1}{3}} (1+b^3+1)^{\frac{1}{3}}
(1+1+c^{3})^{\frac{1}{3}} \geq (a+b+c). (1) \) Dar pentru
orice numere pozitive \( a \), expresiile \( a^2-1 \) si \( a^3-1 \) au acelasi semn. Deci \( 0 \leq (a^3-1)(a^2-1) = a^5-a^3-a^2+1 \) de unde
\( a^5-a^2+3 \geq a^3+2. \) In consecinta
\( (a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3) \geq (a^3+2)(b^3+2)(c^3+2) \) si tinand cont de (1) rezulta inegalitatea din enunt.

4. [Japonia]. Fie \( a,b,c \) numere reale pozitive
care satisfac \( a^2 \leq b^2+c^2, b^2 \leq a^2+c^2, c^2 \leq a^2+b^2 \). Sa se arate ca \( (a+b+c)(a^2+b^2+c^2)(a^3+b^3+c^3) \geq
4(a^6+b^6+c^6). \) si sa se determine cand are loc egalitatea.

Solutie: In inegalitatea (2) facem substitutia \( a_{1}
= a^{\frac{1}{3}}, a_{2} = b^{\frac{1}{3}}, a_{3} =
c^{\frac{1}{3}}, b_{1} = a^{\frac{2}{3}}, b_{2} =
b^{\frac{2}{3}}, b_{3} = c^{\frac{2}{3}}, c_{1} = a, c_{2} =
b, c_{3} = c \). Obtinem \( (a+b+c)(a^2+b^2+c^2)(a^3+b^3+c^3) \geq
(a^2+b^2+c^2)^3, \) si este suficient sa demonstram ca
\( (a^2+b^2+c^2)^3 \geq 4(a^6+b^6+c^6) \). Ridicand la putere si
grupand termenii, ramane sa demonstram ca
\( a^4(a^2+c^2)+b^4(a^2+c^2)+c^4(a^2+b^2)+2a^2b^2c^2 \geq
a^6+b^6+c^6, \) inegalitate ce rezulta imediat aplicand ipoteza.
Evident ca egalitatea are loc daca si numai daca \( a=b=c=0 \).

5. [APMO] Fie \( a,b,c \) numere reale. Sa se arate
\( \left(1+ \frac{a}{b}\right) \left(1 + \frac{b}{c} \right) \left(
1+ \frac{c}{a} \right) \geq 2 \left(1+ \frac{a+b+c}{\sqrt[3]{abc}}
\right). \)

6. [Putnam - 2003] Fie \( a_{1},a_{2}, \ldots,
a_{n} \) si \( b_{1}, b_{2}, \ldots, b_{n} \) numere reale nenegative. Sa se arate ca: \( (a_{1}a_{2} \cdots a_{n})^{\frac{1}{n}} \leq ((a_{1}+b_{1})(a_{2}+b_{2}) \cdots (a_{n}+b_{n}))^{\frac{1}{n}}. \)

7. [T.Andreescu, G. Dospinescu] (din [1]) Sa se arate ca daca a,b,c sunt numere reale pozitive cu \( a+b+c=1 \) atunci \( (a^2+b^2)(b^2+c^2)(c^2+a^2) \geq 8(a^2b^2+b^2c^2+c^2a^2)^2. \)

8. [G. Dospinescu] Sa se gaseasca minimul expresiei \( \sum_{i=1}^{n} \sqrt{ \frac{a_1 a_2 \cdots a_n}{1-(n-1)a_i}} \) unde \( a_1, a_2, \ldots, a_n < \frac{1}{n-1} \), \( a_1+a_2+ \ldots + a_n = 1 \) si \( n>2 \) intreg.

Bibliografie:

[1.] T. Andreescu, V. Cartoaje, G. Dospinescu si M. Lascu - Old and New Inequalities, GIL Publishing House, 2004

[2.] V. Cartoaje - Algebraic Inequalities. Old and New Methods, GIL Publishing House, 2006.

[3.] P.K.Hung - Secrets in Inequalities, GIL Publishing House, 2007

P.S. Invit cititorii de pe forum sa completeze aceasta incercare de nota.

Mircea Lascu

In memoria domnului profesor Alexandru Lupas.

9. [D.Grinberg, http://www.mathlinks.ro/viewtopic.php?t=6573] Fie \( x, y, z \) numere reale pozitive. Sa se arate ca
\( \frac{\sqrt{y+z}}{x}+\frac{\sqrt{z+x}}{y}+\frac{\sqrt{x+y}}{z}\geq \frac{4\left( x+y+z\right) }{\sqrt{\left(y+z\right) \left(z+x\right) \left(x+y\right) }}. \)
Solutie (C. Pohoata)
Din inegalitatea lui Holder avem
\( \left(\frac{\sqrt{y+z}}{x}+\frac{\sqrt{z+x}}{y}+\frac{\sqrt{x+y}}{z}\right)^{2} \left((y+z)^{2}x^{2}+(z+x)^{2}y^{2}+(x+y)^{2}z^{2}\right) \geq 8(x+y+z)^{3}. \)
Deci ne ramane sa demonstram ca:
\( \frac{4(x+y+z)^{3}}{x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}+x^{2}yz+xy^{2}z+xyz^{2}} \geq \frac{16(x+y+z)^{2}}{(x+y)(y+z)(z+x)}, \)
ceea ce se reduce la a arata ca
\( \left(\sum_{cyc}{x}\right) \left(\sum_{cyc}{x^{2}y}+\sum_{cyc}{xy^{2}}+2xyz\right) \geq 4\sum_{cyc}{x^{2}y^{2}}+4\sum_{cyc}{x^{2}yz}, \)
evident adevarata, fiind echivalenta cu
\( \sum_{cyc}{xy(x^{2}+y^{2})} \geq 2\sum_{cyc}{x^{2}y^{2}} \) la randul ei adevarata conform inegalitatii \( x^{2}+y^{2} \geq 2xy \) si analoagele.

10. [http://www.mathlinks.ro/viewtopic.php?t=167079] Fie \( a, b, c \) numere reale pozitive. Sa se arate ca
\( \sum_{cyc}\frac{a}{\sqrt{ab+b^{2}}}\geq\frac{3}{\sqrt{2}} \)

Solutie (C. Pohoata) Consideram \( a=x^{2}, b=y^{2}, c=z^{2}. \)Din inegalitatea lui Holder avem:
\( \left(\sum_{cyc}{\frac{x^{2}}{y\sqrt{x^{2}+y^{2}}}}\right)\left(\sum_{cyc}{\sqrt{x^{2}+y^{2}}}\right)\left(\sum_{cyc}{xy}\right)\geq (x+y+z)^{3}. \)
Deci ne ramane sa demonstram ca
\( 2\left(x+y+z\right)^{6}\geq 9\left(\sum_{cyc}{xy}\right)^{2}\left(\sum_{cyc}{\sqrt{x^{2}+y^{2}}}\right)^{2}. \)
Dar pe de alta parte din inegalitatea Cauchy-Schwarz
\( \sum_{cyc}{\sqrt{x^{2}+y^{2}}}\leq\sqrt{6(x^{2}+y^{2}+z^{2})} \)
Deci, problema se reduce la a arata ca
\(
\left(x+y+z\right)^{6}\geq 27\left(xy+yz+zx\right)^{2}\left(x^{2}+y^{2}+z^{2}\right), \)
care este adevarata si bine-cunoscuta, apartinandu-i lui Michael Rozenberg. (A se vedea http://www.mathlinks.ro/viewtopic.php?t=163253)

11. [Michael Rozenberg, http://www.mathlinks.ro/viewtopic.php?t=163253] Fie \( a, b, c \) numere reale pozitive satisfacand \( a^{4}+b^{4}+c^{4}=3 \). Sa se arate ca
\( \frac{a^{2}}{2b+c}+\frac{b^{2}}{2c+a}+\frac{c^{2}}{2a+b}\geq 1. \)

12. [Ukraine 2007] Fie \( a, b, c \)numere reale pozitive. Sa se arate ca
\( \frac{1+a^{2}}{\sqrt{2a^{2}+3ab-c^{2}}}+\frac{1+b^{2}}{\sqrt{2b^{2}+3bc-a^{2}}}+\frac{1+c^{2}}{\sqrt{2c^{2}+3ca-b^{2}}}\ge2(a+b+c). \)

13. [W.Janous] Fie \( a, b, c \)numere reale pozitive. Sa se arate ca

\( \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}} \)

14. [Vo Quoc Ba Can, Mathematical Reflections, 2/2007] Fie\( a, b, c \) numere reale pozitive. Sa se arate ca
\( \sum_{cyc}\sqrt{\frac{b+c}{a}}\ge\sqrt{\frac{16(a+b+c)^{3}}{3(a+b)(b+c)(c+a)}}. \)

15. [Vasile Cartoaje, http://www.mathlinks.ro/viewtopic.php?t=98994] Fie \( a, b, c \) numere reale pozitive satisfacand \( a^{2}+b^{2}+c^{2}=a+b+c \). Sa se arate ca
\( ab+bc+ca \geq a^2b^2+b^2c^2+c^2a^2. \)

16. [http://www.mathlinks.ro/viewtopic.php?t=147675] Fie \( a, b, c \) numere reale pozitive si k un numar real mai mare sau egal cu \( -2. \) Sa se arate ca
\( \sqrt{\frac{a^{2}}{a^{2}+kab+b^{2}}}+\sqrt{\frac{b^{2}}{b^{2}+kbc+c^{2}}}+\sqrt{\frac{c^{2}}{c^{2}+kca+a^{2}}}\ge \min \left(1;\frac{3}{\sqrt{k+2}}\right) \)

Cosmin Pohoata wrote: 15. [Vasile Cartoaje, http://www.mathlinks.ro/viewtopic.php?t=98994] Fie \( a, b, c \) numere reale pozitive satisfacand \( a^{2}+b^{2}+c^{2}=a+b+c \). Sa se arate ca
\( ab+bc+ca \geq a^2b^2+b^2c^2+c^2a^2. \)

Solutie.

Deoarece \( \sum ab=\left(\frac{a+b+c}{a^{2}+b^{2}+c^{2}}\right)^{2}\left(\sum ab\right) \) si \( \sum a^{2}b^{2}=\left(\frac{a+b+c}{a^{2}+b^{2}+c^{2}}\right)^{4}\left(\sum a^{2}b^{2}\right) \) avem ca \( \sum ab\ge\sum a^{2}b^{2}\Longleftrightarrow\left(\sum a^{2}\right)^{3}\ge\left(\sum a\right)^{2}\left(\sum a^{2}b^{2}\right) \), care se obtine inmultind inegalitatile \( \sum a^{2}\ge\frac{1}{3}\left(\sum a\right)^{2},\ \left(\sum a^{2}\right)^{2}\ge3\sum a^{2}b^{2} \).

17. http://www.mathlinks.ro/viewtopic.php?t=312818

18. http://www.mathlinks.ro/viewtopic.php?p=1686316#1686316