Inegalitate 1

[Marius Mainea]

Daca a,b,c>0, sa se arate ca

\( \frac{(b+c)(b^2+c^2)}{a}+\frac{(c+a)(c^2+a^2)}{b}+\frac{(a+b)(a^2+b^2)}{c}\ge 4(a^2+b^2+c^2). \)

Gheorghe Szollosy, RMT/4/2008

[Claudiu Mindrila]

Avem:
\( \sum \frac{(b+c)(b^2+c^2)}{a} \geq \sum \frac{2bc(b+c)}{a}= \frac{b^2c}{a}+\frac{bc^2}{a}+\frac{c^2a}{b}+\frac{ca^2}{b}+\frac{a^2b}{c}+\frac{ab^2}{c}=\left(\frac{b^2c}{a}+\frac{ab^2}{c} \right)+ \left(\frac{bc^2}{a}+\frac{c^2a}{b} \right) + \left( \frac{ca^2}{b}+ \frac{a^2b}{c} \right) \geq 4(a^2+b^2+c^2) \).
Am folosit doar inegalitatea dintre media aritmetica si media geometrica.

[Marius Mainea]

Avem:
\( \sum \frac{(b+c)(b^2+c^2)}{a} \geq \sum \frac{2bc(b+c)}{a}= 2(\frac{b^2c}{a}+\frac{bc^2}{a}+\frac{c^2a}{b}+\frac{ca^2}{b}+\frac{a^2b}{c}+\frac{ab^2}{c})=2(\left(\frac{b^2c}{a}+\frac{ab^2}{c} \right)+ \left(\frac{bc^2}{a}+\frac{c^2a}{b} \right) + \left( \frac{ca^2}{b}+ \frac{a^2b}{c} \right) )\geq 4(a^2+b^2+c^2) \).
Am folosit doar inegalitatea dintre media aritmetica si media geometrica.

[Claudiu Mindrila]

Solutia 2. Inmultim cu \( 2 \) inegalitatea ceruta si obtinem ca
\( \sum \frac{2(b+c)(b^2+c^2}{a} \geq \frac{(b+c)^2 \cdot (b+c)}{a} \geq 8(a^2+b^2+c^2) \), aceasta inegalitate fiind demonstrata AICI .