bất đẳng thức hay

C

chaizo1234567

đề thế này phải o bạn

$\frac{a^2014+b^2014+c^2014}{a^2013+b^2013+c^2013}$\geq$\frac{a+b+c}{3}$
 
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C

conga222222

$\begin{array}{l}
chuan\;hoa\;a + b + c = 3\;\left( {tu\;tim\;hieu\;cach\;chuan\;hoa\;nhe} \right)\\
\to can\;chung\;minh:\;{a^{2014}} + {b^{2014}} + {c^{2014}} \ge {a^{2013}} + {b^{2013}} + {c^{2013}}\\
\cos i:\\
2013{a^{2014}} + 1 = \underbrace {{a^{2014}} + ... + {a^{2014}}}_{2013\;so} + 1 \ge 2014\sqrt[{2014}]{{{a^{2014*2013}}}} = 2014{a^{2013}}\\
{a^{2014}} + 2013 = {a^{2014}} + \underbrace {1 + ... + 1}_{2013\;so} \ge 2014\sqrt[{2014}]{{{a^{2014}}}} = 2014a\\
\to 2014{a^{2014}} + 2014 \ge 2014a + 2014{a^{2013}}\\
\leftrightarrow {a^{2014}} + 1 \ge a + {a^{2013}}\\
tuong\;tu\\
{b^{2014}} + 1 \ge b + {b^{2013}}\\
{c^{2014}} + 1 \ge c + {c^{2013}}\\
\to {a^{2014}} + {b^{2014}} + {c^{2014}} + 3 \ge a + b + c + {a^{2013}} + {b^{2013}} + {c^{2013}}\\
\leftrightarrow {a^{2014}} + {b^{2014}} + {c^{2014}} \ge {a^{2013}} + {b^{2013}} + {c^{2013}}\;\left( {do\;a + b + c = 3} \right)
\end{array}$
 
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