giúp đỡ táon 8

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haidang96

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le_tien

[TEX]\frac{a^5}{bc} + \frac{b^5}{ac} + \frac{c^5}{ba} = \frac{1}{abc} (a^6 + b^6 + c^6) \geq [/TEX]
[TEX]\geq \frac{1}{3abc}(a^3 + b^3 + c^3)^2 = \frac{1}{3abc}(a^3 + b^3 + c^3)(a^3 + b^3 + c^3) \geq[/TEX]
[TEX]\geq (a^3 + b^3 + c^3). \frac{1}{3abc}. 3\sqrt[3]{a^3b^3c^3} = a^3 + b^3 + c^3[/TEX]

Dấu [TEX]=[/TEX] xảy ra khi [TEX]a = b = c[/TEX]
 
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