Subiectul 3 OJM IX 2009

[mihai++]

Fie \( a,b\geq0,x,y>0 \).
Demonstrati ca \( \frac{a^3}{x^2}+\frac{b^3}{y^2}\geq\frac{(a+b)^3}{(x+y)^2} \).

[mihai++]

Inegalitatea e echivalenta cu \( (1+\frac{y}{x})^2a^3+(1+\frac{x}{y})^2b^3\geq(a+b)^3 \) si din Cauchy stim:
\( (a+b)((1+\frac{y}{x})^2a^3+(1+\frac{x}{y})^2b^3)\geq(a^2+b^2+a^2\cdot\frac{y}{x}+b^2\cdot\frac{x}{y})^2\geq (a^2+b^2+2\sqrt{a^2b^2\cdot\frac{x}{y}\cdot\frac{y}{x}})^2=(a+b)^4 \) egalitatea se intampla cand \( a=b,x=y \).

[DrAGos Calinescu]

In barem e o rezolvare destul de urata.
Eu am demonstrat folosind binecunoscuta inegalitate \( \frac{x^2}{a}+\frac{y^2}{b}\ge\frac{(x+y)^2}{a+b} \), la care am inmultit cu \( \frac{x+y}{a+b} \)
Avem de demonstrat ca\( \frac{a^2(a+b)}{x(x+y)}+\frac{b^2(a+b)}{y(x+y)}\le\frac{a^3}{x^2}+\frac{b^3}{y^2} \)
\( \Longleftrightarrow \frac{a^3y^2+b^3x^2}{x^2y^2}\ge\frac{a^2x(a+b)+b^2y(a+b)}{xy(x+y)} \)
\( \Longleftrightarrow a^3y^2(x+y)+b^3x^2(x+y)\ge a^2x^2y(a+b)+b^2xy^2(a+b) \)
\( \Longleftrightarrow a^3xy^2+a^3y^3+b^3x^3+b^3x^2y\ge a^3x^2y+a^2bx^2y+b^2xay^2+b^3xy^2 \)
\( \Longleftrightarrow a^3y^3+b^3x^3\ge a^2bx^2y+ab^2xy^2 \)
Impartim prin \( abxy \), iar relatia devine echivalenta cu
\( \frac{a^2x^2}{bx}+\frac{b^2x^2}{ax}\ge ax+by \), evidenta folosind inegalitatea cu care am inceput.
E destul de lunga, dar asa mi-a iesit

[Al3xx]

A fost o problema destul de ciudata, pentru ca nu se putea aplica DIRECT alta inegalitate cunoscuta... trebuia sa prelucrezi putin.

Eu am incercat sa inlocuiesc in \( \frac{a^2}{x} + \frac{b^2}{y} \ge \frac{(a+b)^2}{x+y} \) pe \( a \) cu \( \sqrt{a^3} \) pe b cu \( \sqrt{b^3} \), pe \( x \) cu \( x^2 \) si pe \( y \) cu \( y^2 \) si sa compar fractiile formate, dar nu da...

[Marius Mainea]

Solutie ,,rapida'':

Folosim inegalitatea lui Holder:

\( (a_1b_1c_1+x_1y_1z_1)^3\le(a_1^3+x_1^3)(b_1^3+y_1^3)(c_1^3+z_1^3) \) pentru orice numere nenegative.

Asadar \( (a+b)^3=(\frac{a}{\sqrt[3]{x^2}}\cdot \sqrt[3]{x}\cdot \sqrt[3]{x}+\frac{b}{\sqrt[3]{y^2}}\cdot \sqrt[3]{y}\cdot \sqrt[3]{xy})^3\le(\frac{a^3}{x^2}+\frac{b^3}{y^2})(x+y)(x+y) \)

care este echivalenta cu inegalitatea din enunt.

[Virgil Nicula]

Fie numerele pozitive \( a \) , \( b \) , \( x \) , \( y \) . Aratati ca \( \frac{a^3}{x^2}\ +\ \frac{b^3}{y^2}\ \ge\ \frac{(a+b)^3}{(x+y)^2} \) .

Demonstratie. Nu restrangem generalitatea daca presupunem \( \underline{\overline{\left\|\ b=y=1\ \right\|}} \) . Inegalitatea propusa devine :

\( \left(a^3+x^2\right)(x+1)^2\ \ge\ x^2(a+1)^3 \) . Notam \( \underline{\overline{\left\|\ a=xt\ ,\ t\ >\ 0\ \right\|}} \) . Inegalitatea precedenta devine :

\( \left(xt^3+1\right)(x+1)^2\ \ge\ (xt+1)^3\ \Longleftrightarrow\ \) \( x\left(2t^3-3t^2+1\right)+\left(t^3-3t+2\right)\ \ge\ 0\ \Longleftrightarrow \)

\( (t-1)^2\cdot\left[x(2t+1)+(t+2)\right]\ \ge\ 0 \) . Avem egalitate daca si numai daca \( t=1 \) , adica \( a=x \) .

Observatie. Vom arata ca \( \underline{\overline{\left\|\ \frac{a^3}{x^2}\ +\ \frac{b^3}{y^2}\ \ge\ \frac{(a+b)^3}{(x+y)^2}\ \Longleftrightarrow\ (bx-ay)^2\cdot\left[bx^2+2(a+b)xy+ay^2\right]\ \ge \ 0\ \right\|}}\ . \)

Revenind la inegalitatea propusa, ea este echivalenta cu \( \left(\frac ax-\frac by\right)^2\cdot\left[x\left(\frac {2a}{x}+\frac by\right)+\left(\frac ax+\frac {2b}{y}\right)\right]\ \ge\ 0\ \Longleftrightarrow \)

\( (bx-ay)^2\cdot\left[bx^2+2(a+b)xy+ay^2\right]\ \ge \ 0 \) cu egalitate daca si numai daca \( \frac ax=\frac by \) .

Generalizare. \( \{\ a\ ,\ b\ ,\ x\ ,\ y\ \}\subset \mathbb R^*_+\ \ \wedge\ \ \{\ m\ ,\ n\ \}\subset \mathbb N^* \) \( \Longrightarrow\ \frac{a^m}{x^n}\ +\ \frac{b^m}{y^n}\ \ge\ \frac{\left(a^{\frac {m}{n+1}}+b^{\frac {m}{n+1}}\right)^{n+1}}{(x+y)^n}\ . \)

Aplicam ineg. Huygens : \( (x+y)^n\left(a^my^n+b^mx^n\right)\ \ge\ \left(\sqrt[n+1]{a^mx^ny^n}+\sqrt[n+1]{b^mx^ny^n}\right)^{n+1}\ . \) Deci

\( (x+y)^n\left(a^my^n+b^mx^n\right)\ \ge\ x^ny^n\left(a^{\frac {m}{n+1}}+b^{\frac {m}{n+1}}\right)^{n+1}\ \Longleftrightarrow\ \frac{a^m}{x^n}\ +\ \frac{b^m}{y^n}\ \ge\ \frac{\left(a^{\frac {m}{n+1}}+b^{\frac {m}{n+1}}\right)^{n+1}}{(x+y)^n}\ . \)

[Virgil Nicula]

... inegalitatea Holder \( (a_1b_1c_1+x_1y_1z_1)^3\le(a_1^3+x_1^3)(b_1^3+y_1^3)(c_1^3+z_1^3) \) peste \( \mathbb R_+\ . \)

Inegalitatea "tare" mentionata de prof. Marius Mainea se numeste inegalitatea Huygens, dupa cate stiu eu.

Ineg. Huygens : \( n\in\mathbb{N}^*\ ,\ a_k>0\ ,\ b_k>0\ ,\ (\forall )\ k\in\overline{1,n} \) \( \ \Longrightarrow\ \left(\prod_{k=1}^na_k+\prod_{k=1}^nb_k\right)^n\ \le\ \prod_{k=1}^n\left(a_k^n+b_k^n\right)\ . \)

O forma echivalenta a acestei inegalitati este \( \left(\sqrt[n]{\prod_{k=1}^na_k}+\sqrt[n]{\prod_{k=1}^nb_k}\right)^n\ \le\ \prod_{k=1}^n\left(a_k+b_k\right)\ . \)

Ineg. Holder : \( \left\|\ \begin{array}{c}
n\in\mathbb{N}^*\ ,\ p>1\ ,\ \frac 1p+\frac 1q=1\\\\\\\\
a_k>0\ ,\ b_k>0\ ,\ (\forall )\ k\in\overline{1,n}\end{array}\ \right\| \) \( \Longrightarrow \) \( \sum_{k=1}^na_kb_k\ \le\ \left(\sum_{k=1}^na_k^p\right)^{\frac 1p}\cdot\left(\sum_{k=1}^nb_k^q\right)^{\frac 1q}\ . \)

[Mihai Berbec]

Ultima inegalitate seamana cu inegalitatea lui Holder : \( ||fg||_1\le || f||_p\cdot || g||_q \), unde p si q sunt conjugate.

[Cezar Lupu]

Pai este chiar inegalitatea lui Holder in discret.

[zeta]

\( \frac{x}{x+y}(\frac{a}{x})^3+\frac{y}{x+y}(\frac{b}{y})^3\geq(\frac{x}{x+y}\frac{a}{x}+\frac{y}{x+y}\frac{b}{y})^3 \)
de unde rezulta ineg din enuntz.