Problema 3, TST OBMJ 2006

[Claudiu Mindrila]

Fie \( a,b,c \) numere reale strict pozitive cu suma \( 1. \) Sa se arate ca \( \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq 3(a^2+b^2+c^2) \)
Mircea Lascu, TST OBMJ, 2006

[mihai++]

Sumele de mai jos sunt ciclice:
\( a+b+c=1\Rightarrow \sum \frac{a^2}{b}\geq3\sum a^2\Leftrightarrow \sum a\cdot \sum \frac{a^2}{b}\geq3\sum a^2\Leftrightarrow \sum \frac{a^3}{b}+\sum \frac{ab^2}{c}\geq 2\sum a^2 \).
Aplicam Cauchy:
\( (ab+bc+ca+ac+ba+cb)(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+\frac{ab^2}{c}+\frac{bc^2}{a}+\frac{ca^2}{b})\geq(a^2+b^2+c^2+ab+bc+ca)^2\Rightarrow\\ \sum \frac{a^3}{b}+\sum\frac{ab^2}{c}\geq\frac{(a^2+b^2+c^2+ab+bc+ca)^2}{2(ab+bc+ca)}\geq\frac{4(a^2+b^2+c^2)(ab+bc+ca)}{2(ab+bc+ca)}=2(a^2+b^2+c^2)\\ \).
Egalitate la \( a=b=c=\frac{1}{3} \)

[Claudiu Mindrila]

Cu inegalitatea Cauchy-Buniakowski-Schwarz avem:
\( \sum\frac{a^{2}}{b}=\sum\frac{a^{4}}{a^{2}b}\ge\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a^{2}b+b^{2}c+c^{2}a} \).

Problema revine la a arata ca \( \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a^{2}b+b^{2}c+c^{2}a}\ge3\left(a^{2}+b^{2}+c^{2}\right)\Longleftrightarrow a^{2}+b^{2}+c^{2}\ge3\left(a^{2}b+b^{2}c+c^{2}a\right)\Longleftrightarrow\left(a^{2}+b^{2}+c^{2}\right)\left(a+b+c\right)\ge3\left(a^{2}b+b^{2}c+c^{2}a\right) \).

Aceasta ultima inegalitate este adevarata, ea fiind echivalenta cu \( a\left(a-b\right)^{2}+b\left(b-c\right)^{2}+c\left(c-a\right)^{2}\ge0 \).