Inegalitate din SHL 2004

[Claudiu Mindrila]

Fie \( a,\ b,\ c \) numere reale strict pozitive a. i. \( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \).
Sa se demonstreze inegalitatea

\( \sum\frac{a}{bc\left(b+c\right)} \ge \frac{3}{2} \)

Cezar si Tudorel Lupu, lista scurta, 2004

[Mateescu Constantin]

Inegalitatea \( \begin{array}{cccc} \frac 1a & + & \frac 1b & + & \frac 1c & \ge & \frac ab & + & \frac bc & + & \frac ca\end{array} \) este echivalenta cu : \( ab+bc+ca\ \ge\ a^2c+b^2a+c^2b\ \ (\ast) \)

Aplicam inegalitatea C.B.S : \( (ac+ba+cb)^2=\(\sqrt{a^2c}\cdot\sqrt c+\sqrt{b^2a}\cdot\sqrt a+\sqrt{c^2b}\cdot\sqrt b\)^2\ \stackrel{\small CBS}{\le}\ (a^2c+b^2a+c^2b)(a+b+c) \)

\( \Longrightarrow\ (ab+bc+ca)^2\le (a^2c+b^2a+c^2b)(a+b+c)\ \le^{(\ast)}\ (ab+bc+ca)(a+b+c)\ \Longrightarrow\ ab+bc+ca\le a+b+c\ (1) \)

Din inegalitatea cunoscuta \( \sqrt{3abc(a+b+c)}\ \le\ ab+bc+ca \) si din \( (1) \) obtinem : \( a+b+c\ \ge\ \sqrt{3abc(a+b+c)}\ \Longleftrightarrow \)

\( \Longleftrightarrow\ \fbox{\ a+b+c\ \ge\ 3abc\ }\ (\ast\ast) \) . Acum, pentru inegalitatea de demonstrat aplicam C.B.S si tinem cont si de\( \ (\ast\ast)\ : \)

Intr-adevar , \( \sum\ \frac{a}{bc(b+c)}=\frac{1}{abc}\sum\ \frac{a^2}{b+c}\ \ge^{\small CBS}\ \frac{1}{abc}\ \cdot\ \frac{(a+b+c)^2}{2(a+b+c)}=\frac{(a+b+c)}{2abc}\ \ge^{(\ast\ast)}\ \frac 32\ \ ,\ \ \mbox\normal{O.K.} \)

[Mateescu Constantin]

Cu aceeasi conditie se poate demonstra si inegalitatea mai tare : \( \fbox{\ \sum\ \frac{a}{bc(b+c)}\ \ge\ \frac{9(a^2+b^2+c^2)}{2(a+b+c)^2}\ }\ \ge\ \frac 32 \) .