Inegalitate cu radicali

[Marius Mainea]

Sa se arate ca daca a,b,c>0 atunci :

\( \sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}\ge \sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}. \)

[Claudiu Mindrila]

Fie \( a=x^3,b=y^3,c=z^3 \) cu \( x,y,z>0 \). Inegalitatea ceruta devine: \( 3(x^3+y^3+z^3)\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right) \geq \left( \frac{x}{y}+\frac{y}{z}+\frac{z}{x} \right) ^3 \). Aceasta din urma rezulta imediat din inegalitatea Holder.

[Marius Mainea]

Fie \( a=x^3,b=y^3,c=z^3 \) cu \( x,y,z>0 \). Inegalitatea ceruta devine: \( 3(x^3+y^3+z^3)\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right) \geq \left( \frac{x}{y}+\frac{y}{z}+\frac{z}{x} \right) ^3 \). Aceasta din urma rezulta imediat din inegalitatea Holder.

Ar fi bine daca ai enunta inegalitatea lui Holder.

[Claudiu Mindrila]

Pentru orice \( a_{1},a_{2},a_{3},b_{1},b_{2},b_{3},c_{1},c_{2},c_{3}>0 \) avem: \( \left(a_{1}^{3}+a_{2}^{3}+a_{3}^{3}\right)\left(b_{1}^{3}+b_{2}^{3}+b_{3}^{3}\right)\left(c_{1}^{3}+c_{2}^{3}+c_{3}^{3}\right)\geq\left(a_{1}b_{1}c_{1}+a_{2}b_{2}c_{2}+a_{3}b_{3}c^{3}\right) \).

Revenind la problema,
\( \left(1^{3}+1^{3}+1^{3}\right)\left(x^{3}+y^{3}+z^{3}\right)\left(\frac{1}{y^{3}}+\frac{1}{z^{3}}+\frac{1}{x^{3}}\right)\geq\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)^{3} \).

[Marius Mainea]

Asa mai merge .

Gandeste-te acum la o demonstratie fara Holder, sau demonstreaza-le pe celelalte.

[Marius Mainea]

Prin ridicare la cub si desfacerea parantezelor inegalitatea devine:


\( 9+3\sum{(\frac{a}{b}+\frac{b}{a})}\ge\sum{\frac{a}{b}}+6+3\sum{(\sqrt[3]{\frac{a^2}{bc}}+\sqrt[3]{\frac{bc}{a^2}})} \) sau

\( 3+2\sum{\frac{a}{b}}+3\sum{\frac{b}{a}}\ge 3\sum{(\sqrt[3]{\frac{a^2}{bc}}+\sqrt[3]{\frac{bc}{a^2}})} \)

care este adevarata (demonstrati!).