Chủ đề này có 3 trả lời

[Ngoc Hung]

Cho a, b, c > 0. CMR $\frac{a^{4}}{b^{3}(c+2a)}+\frac{b^{4}}{c^{3}(a+2b)}+\frac{c^{4}}{a^{3}(b+2c)}\geq 1$

[canhhoang30011999]

Cho a, b, c > 0. CMR $\frac{a^{4}}{b^{3}(c+2a)}+\frac{b^{4}}{c^{3}(a+2b)}+\frac{c^{4}}{a^{3}(b+2c)}\geq 1$

ta có

$\sum \frac{a^{4}}{b^{3}(c+2a)}= \sum \frac{\frac{a^{4}}{b^{2}}}{b(c+2a)}$

$\geq \frac{(\sum \frac{a^{2}}{b})}{3(ab+bc+ca)}$

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

[I Am Gifted So Are You]

Áp dụng Am-gm ta có
$\frac{a^4}{b^3(c+2a)}+\frac{c+2a}{9a}+\frac{1}{3}\geq \frac{a}{b}$
$\frac{b^4}{c^3(a+2b)}+\frac{a+2b}{9b}+\frac{1}{3}\geq \frac{b}{c}$
$\frac{c^4}{a^3(b+2c)}+\frac{b+2c}{9c}+\frac{1}{3}\geq \frac{c}{a}$
Cộng vế vs vế
$\Rightarrow VT\geq \frac{8}{9}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})-\frac{5}{3}\geq 1(dpcm)$

[nguyenhongsonk612]

Cho a, b, c > 0. CMR $P= \frac{a^{4}}{b^{3}(c+2a)}+\frac{b^{4}}{c^{3}(a+2b)}+\frac{c^{4}}{a^{3}(b+2c)}\geq 1$

Một cách khác, không dùng S-vác.

Giải:

Ta có BĐT phụ sau: $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq a+b+c$

Sử dụng BĐT Bunhiacopxki, ta có:

$(a+b+c)^2\leq (\sum \frac{a^2}{b})^2= (\frac{a^2}{b\sqrt{b(c+2a)}}.\sqrt{b(c+2a)}+\frac{b^2}{c\sqrt{c(a+2b)}}.\sqrt{c(a+2b)}+\frac{c^2}{a\sqrt{a(b+2c)}}.\sqrt{a(b+2c)})^2\leq P.3(ab+bc+ca)\leq (a+b+c)^2.P$

$\Rightarrow P\geq 1$

BĐT được chứng minh.

P/s: Có lẽ dùng S-vác sẽ nhanh hơn.