Chủ đề này có 3 trả lời

[Vu Thuy Linh]

Cho a, b, c > 0. Cm:

$\frac{a^{3}}{2b+3c}+\frac{b^{3}}{2c+3a}+\frac{c^{3}}{2a+3b}\geq \frac{1}{5}(a^{2}+b^{2}+c^{2})$

[doanlemanhtung191199]

Ta có :$\sum \frac{a^3}{2b+3c}=\sum \frac{a^4}{2ab+3ac}\geq \frac{(a^2+b^2+c^2)^2}{5(ab+bc+ca)}$

Lại có : $a^2+b^2+c^2\geq ab+bc+ca$

Nên: $\sum \frac{a^3}{2b+3c}\geq \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)}=\frac{1}{5}(a^2+b^2+c^2)$

[thuan192]

Cho a, b, c > 0. Cm:

$\frac{a^{3}}{2b+3c}+\frac{b^{3}}{2c+3a}+\frac{c^{3}}{2a+3b}\geq \frac{1}{5}(a^{2}+b^{2}+c^{2})$

$\sum \frac{a^{3}}{2b+3c}=\sum \frac{a^{4}}{2ab+3ac}\geq \frac{\left ( a^{2}+b^{2}+c^{2} \right )^{2}}{5\left ( ab+bc+ca \right )}\geq \frac{1}{5}\left ( a^{2}+b^{2}+c^{2} \right )$                    (  Cauchy-Swcharz)

[Hoang Tung 126]

Theo Cauchy-Swtach có :$\sum \frac{a^3}{2b+3c}=\sum \frac{a^4}{2ab+3ac}\geq \frac{(\sum a^2)^2}{5\sum ab}\geq \frac{(\sum a^2)^2}{5\sum a^2}=\frac{\sum a^2}{5}$