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想請教一題不等式

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想請教一題不等式

朋友問的,卡很久了....希望各位能幫個忙.....冏
設a、b、c、d為正實數,且a^2+b^2+c^2+d^2=4
求證:(a^2/b)+(b^2/c)+(c^2/d)+(d^2/a)>=4

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回復 1# eyeready 的帖子

\(\left( \frac{{{a}^{2}}}{b}+\frac{{{b}^{2}}}{c}+\frac{{{c}^{2}}}{d}+\frac{{{d}^{2}}}{a} \right)\left( {{a}^{2}}b+{{b}^{2}}c+{{c}^{2}}d+{{d}^{2}}a \right)\ge {{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)}^{2}}=16\)

原題即證明\({{a}^{2}}b+{{b}^{2}}c+{{c}^{2}}d+{{d}^{2}}a\le 4\)

\(\begin{align}
  & {{\left( {{a}^{2}}b+{{b}^{2}}c+{{c}^{2}}d+{{d}^{2}}a \right)}^{2}} \\
& \le \left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)\left( {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{d}^{2}}+{{d}^{2}}{{a}^{2}} \right) \\
& =4\left( {{a}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{d}^{2}} \right) \\
& \le 4\times \frac{{{\left[ \left( {{a}^{2}}+{{c}^{2}} \right)+\left( {{b}^{2}}+{{d}^{2}} \right) \right]}^{2}}}{4} \\
& =16 \\
&  \\
& {{a}^{2}}b+{{b}^{2}}c+{{c}^{2}}d+{{d}^{2}}a\le 4 \\
\end{align}\)

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回復 2# thepiano 的帖子

非常感謝...太強了!!

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