278. In memoria profesorului Alexandru Lupas.

by Virgil Nicula, May 20, 2011, 3:18 AM

In memoria domnului profesor Alexandru Lupas.
Mircea Lascu wrote:
Inegalitatea lui Holder. Daca $a_{1}\ ,\ a_{2},\ \ldots,\ a_{n}\ ,$ $b_{1}\ ,\ b_{2}\ ,\ \ldots, b_{n}$ si $p,q$ sunt 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 $\frac{x^{p}}{p}+\frac{y^{q}}{q}+\frac{z^{r}}{r} \geq xyz\ .$

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


Demonstratie. 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}\ ,$ rezulta inegalitatea.


3. T. Andreescu, USAMO 2004 Fie numerele reale pozitive $a,b,c\ .$ Sa se arate ca $(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3) \geq (a+b+c)^{3}\ .$

Demonstratie. Din 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 numar pozitiv $a$ , 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

$(a+b+c)(a^2+b^2+c^2)(a^3+b^3+c^3) \geq$ $(a^6+b^6+c^6)\ .$ si sa se determine cand are loc egalitatea.


Demonstratie. 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\implies$ $(a+b+c)(a^2+b^2+c^2)(a^3+b^3+c^3) \geq$

$(a^2+b^2+c^2)^3$ si este suficient sa demonstram ca $(a^2+b^2+c^2)^3 \geq $ $4(a^6+b^6+c^6)\ .$ Ridicam la putere si grupam termenii. Inegalitatea $a^4(a^2+c^2)+b^4(a^2+c^2)+$

$c^4(a^2+b^2)+2a^2b^2c^2 \geq$ $a^6+b^6+c^6$ 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 $:\ (a_{1}a_{2} \cdots a_{n})^{\frac{1}{n}} \leq \left[(a_{1}+b_{1})(a_{2}+b_{2}) \cdots (a_{n}+b_{n})\right]^{\frac{1}{n}}\ .$

7. T.Andreescu, G. Dospinescu. 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 $\{a_1\ ,\ a_ ,\ \ldots\ ,\ a_n\}\subset\left(0\ ,\ \frac{1}{n-1}\right)\ ,\ a_1+a_2+ \ldots + a_n = 1$ 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
Cosmin Pohoata wrote:
9. D.Grinberg (aici). Fie $x\ ,\ y\ ,\  z$ numere reale pozitive. Sa se arate ca $\frac{\sqrt{y+z}}{x}+\frac{\sqrt{z+x}}{y}+\frac{\sqrt{x+y}}{z}\geq \frac{4\left( x+y+z\right) }{\sqrt{\left(y+z\right) \left(z+x\right) \left(x+y\right) }}\ .$

Demonstratie. 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. (aici). Fie $a\ ,\ b\ ,\ c$ numere reale pozitive. Sa se arate ca $\sum_{cyc}\frac{a}{\sqrt{ab+b^{2}}}\geq\frac{3}{\sqrt{2}}\ .$

Demonstratie. 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 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 binecunoscuta, apartinandu-i lui Michael Rozenberg. (vezi aici).


11. Michael Rozenberg. (aici). 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. Fie numere reale positive $a, b, c$ care satisfac relatia $a^{2}+b^{2}+c^{2}=a+b+c\ .$ Sa se arate ca $ab+bc+ca \geq a^2b^2+b^2c^2+c^2a^2\ .$

Demonstratie. 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 $\sum a^{2}\ge\frac{1}{3}\left(\sum a\right)^{2}$ , $\left(\sum a^{2}\right)^{2}\ge3\sum a^{2}b^{2}\ .$


16. (aici). Fie $a\ ,\  b\ ,\ c$ numere reale pozitive si $k$ un numar real mai mare

sau egal cu $-2\ .$ Sa se arate ca $\sqrt{\frac{a^{2}}{a^{2}+kab+b^{2}}}+\sqrt{\frac{b^{2}}{b^{2}+kbc+c^{2}}}+\sqrt{\frac{c^{2}}{c^{2}+kca+a^{2}}}\ge \min \left(1;\frac{3}{\sqrt{k+2}}\right)\ .$


This post has been edited 78 times. Last edited by Virgil Nicula, May 30, 2016, 8:34 PM

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