Gtnn

C

conga222222

$\eqalign{
& \operatorname{cosi} : \cr
& \frac{1}{{{x^4}}} + \frac{1}{{{y^4}}} + \frac{1}{{{z^4}}} = \underbrace {\frac{1}{{16{x^4}}} + ... + \frac{1}{{16{x^4}}}}_{16\;so} + \underbrace {\frac{1}{{16{y^4}}} + ... + \frac{1}{{16{y^4}}}}_{16\;so} + \frac{1}{{{z^4}}} \geqslant 33\root {33} \of {\frac{1}{{{{16}^{32}}{x^{64}}{y^{64}}{z^4}}}} \cr
& {x^4} + {y^4} + {z^4} = {x^4} + {y^4} + \underbrace {\frac{{{z^4}}}{{16}} + ... + \frac{{{z^4}}}{{16}}}_{16\;so} \geqslant 18\root {18} \of {\frac{{{x^4}{y^4}{z^{64}}}}{{{{16}^{16}}}}} \cr
& \to P = \left( {{x^4} + {y^4} + {z^4}} \right)\left( {\frac{1}{{{x^4}}} + \frac{1}{{{y^4}}} + \frac{1}{{{z^4}}}} \right) \geqslant 594\root {33} \of {\frac{1}{{{{16}^{32}}{x^{64}}{y^{64}}{z^4}}}} *\root {18} \of {\frac{{{x^4}{y^4}{z^{64}}}}{{{{16}^{16}}}}} = 594\root {594} \of {{{\left( {\frac{1}{{{{16}^{32}}{x^{64}}{y^{64}}{z^4}}}} \right)}^{18}}*{{\left( {\frac{{{x^4}{y^4}{z^{64}}}}{{{{16}^{16}}}}} \right)}^{33}}} \cr
& = 594\root {594} \of {\frac{{{z^{2040}}}}{{{{16}^{1104}}{x^{1020}}{y^{1020}}}}} \cr
& ma\;z \geqslant x + y \geqslant 2\sqrt {xy} \to {z^{2040}} \geqslant {2^{2040}}{x^{1020}}{y^{1020}} \cr
& \to P \geqslant 594\root {594} \of {\frac{{{z^{2040}}}}{{{{16}^{1104}}{x^{1020}}{y^{1020}}}}} \geqslant 594\root {594} \of {\frac{{{2^{2040}}{x^{1020}}{y^{1020}}}}{{{{16}^{1104}}{x^{1020}}{y^{1020}}}}} = 594\root {594} \of {\frac{1}{{{2^{2376}}}}} = \frac{{594}}{{{2^4}}} = \frac{{297}}{8} \cr
& dau = \leftrightarrow x = y = \frac{z}{2} \cr} $
 
Q

quanghao98

anh ơi liệu còn cách sử dụng kỹ thuật chọn điểm rơi cauchy schawrz không?em thấy nó ở dạng tổng bình phương nhưng vẫn chưa chứng minh được
 
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