O imbunatatire a unei inegalitati

[Claudiu Mindrila]

In Math. Reflections apare urmatoarea problema:

Daca \( a,\ b,\ c>0 \) atunci \( \frac{a^{3}}{b^{2}+c^{2}}+\frac{b^{3}}{c^{2}+a^{2}}+\frac{c^{3}}{a^{2}+b^{2}}\ge\frac{a+b+c}{2} \)

Va propun o imbunatatire a inegalitatii de mai sus.

Fie \( a, \ b,\ c >0 \). Sa se arate ca \( \frac{a^{3}}{b^{2}+c^{2}}+\frac{b^{3}}{c^{2}+a^{2}}+\frac{c^{3}}{a^{2}+b^{2}}\ge\frac{3}{2}\sqrt{\frac{a^{3}+b^{3}+c^{3}}{a+b+c}} \) .

Claudiu Mindrila

[alex2008]

In Math. Reflections apare urmatoarea problema:

Daca \( a,\ b,\ c>0 \) atunci \( \frac{a^{3}}{b^{2}+c^{2}}+\frac{b^{3}}{c^{2}+a^{2}}+\frac{c^{3}}{a^{2}+b^{2}}\ge\frac{a+b+c}{2} \)

Va propun o imbunatatire a inegalitatii de mai sus.

Fie \( a, \ b,\ c >0 \). Sa se arate ca \( \frac{a^{3}}{b^{2}+c^{2}}+\frac{b^{3}}{c^{2}+a^{2}}+\frac{c^{3}}{a^{2}+b^{2}}\ge\frac{3}{2}\sqrt{\frac{a^{3}+b^{3}+c^{3}}{a+b+c}} \) .

Claudiu Mindrila

Din Cebasev si Cauchy-Schwartz:

\( LHS\ge \frac{1}{3}(a^3+b^3+c^3)\left(\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}\right)\ge \frac{3}{2}\cdot \frac{a^3+b^3+c^3}{a^2+b^2+c^2} \)

Mai ramane de demonstat:

\( \frac{a^3+b^3+c^3}{a^2+b^2+c^2}\ge \sqrt{\frac{a^3+b^3+c^3}{a+b+c}}\Longleftrightarrow (a^3+b^3+c^3)(a+b+c)\ge (a^2+b^2+c^2)^2 \)

adevarata din nou din Cauchy-Schwartz.

Alta imbunatatire:

Fie \( a,b \) si \( c \) trei numere nenegative astfel incat \( ab+bc+ca\neq 0 \). Sa se demonstreze ca:

\( \frac{a^{3}}{a^{2}+b^{2}}+\frac{b^{3}}{b^{2}+c^{2}}+\frac{c^{3}}{c^{2}+a^{2}}\ge\frac{a+b+c}{2} \)