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senconxauxi

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heyday195

Áp dụng BĐT AM-GM :
[TEX]\sqrt{2a( 1 - a^2 )}\leq a^2 + 1 - a^2 = 1[/TEX]
Tg tự: [TEX]\sqrt{2b( 1 - b^2 )}\leq 1[/TEX]
[TEX]\sqrt{2c( 1 - c^2 )}\leq 1[/TEX]
[TEX]\frac { a^2}{\sqrt {1 - a^2}}[/TEX] + [TEX]\frac{ b^2}{\sqrt {1 - b^2}}[/TEX] + [TEX]\frac{ c^2}{\sqrt {1 - c^2}}[/TEX] = [TEX]\frac{2(a^3)}{2a(\sqrt{1 - a^2)}[/TEX] + [TEX]\frac{2(b^3)}{2b(\sqrt{1 - b^2)}[/TEX] + [TEX]\frac{2(c^3)}{2c(\sqrt{1 - c^2)}[/TEX] [TEX]\geq 2( a^3 + b^3 + c^3) = 2[/TEX]
 
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