cho $4(a+b+c)=3abc$
chứng minh $\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\geq \frac{3}{8}$
Ta có $ \frac 34 = \sum \frac 1{ab} \le \frac{(\frac 1a + \frac 1b + \frac 1c)^2}{3}\Rightarrow \frac 1a + \frac 1b + \frac 1c \ge \frac 32$
$VT \ge \frac{1}{9}( \frac 1a + \frac 1b + \frac 1c)^3 \ge \frac 19.(\frac 32)^3= \frac 38$
$\frac 1{a^3} + \frac 1{a^3} + \frac 18 \ge \frac{3}{2a^2}$
$\frac 1{b^3} + \frac 1{b^3} + \frac 18 \ge \frac{3}{2b^2}$
$\frac 1{c^3} + \frac 1{c^3} + \frac 18 \ge \frac{3}{2c^2}$
$\Rightarrow 2VT+ \frac 38 \ge \frac 32 ( \frac 1{a^2} + \frac 1{b^2}+ \frac 1{c^2}) \ge \frac 32( \frac 1{ab} + \frac 1{bc} + \frac 1{ca})= \frac 32. \frac 34$
$\Rightarrow$ đpcm
Áp dụng bđt sau $3(x^3+y^3+z^3)^2 \ge (x^2+y^2+z^2)^3$
Ta có 
$3(\frac 1{a^3}+ \frac 1{b^3} + \frac 1{c^3})^2 \ge (\frac{1}{a^2}+ \frac 1{b^2} + \frac 1{c^2})^3 \ge ( \frac 1{ab} + \frac 1{bc} + \frac 1{ca})^3=(\frac{a+b+c}{abc})^3=(\frac{3}{4})^3$
$\Rightarrow VT \ge \sqrt{\frac{(\frac 34)^3}{3}}= \frac 38$
ak có đk a,b,c> 0 nữa –  nguyenquangtruonghktcute 24-03-16 08:06 PM
mà khoan đề có cho a,b,c >0 ko nhỉ –  tran85295 24-03-16 07:24 PM
chắc ko :v –  tran85295 24-03-16 07:24 PM
thế còn cách nào nữa k :v –  nguyenquangtruonghktcute 24-03-16 06:46 PM
đc chú để tí đăng cho –  tran85295 24-03-16 05:42 PM
Bài này dùng cauchy schwarz đc k nam –  nguyenquangtruonghktcute 24-03-16 05:22 PM

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