toan đại khó dey ba` con 0j

T

thienvamai

cho 3 so dương a,b ,c thoả mãn [TEX]a^2+b^2+c^2=3[/TEX]
chứng minh bất đẳng thức sau;

[TEX]\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq ab+bc+ac[/TEX]

[TEX]2(a^2+b^2+c^2)=a^2+1+b^2+1+c^2+1 \geq 2(a+b+c)[/TEX]
[TEX]\Rightarrow 2(a^2+b^2+c^2)\geq a^2+b^2+c^2 +b+c+a\geq 2(a\sqrt{b}+2b\sqrt{c}+2c\sqrt{a} )[/TEX]
[TEX]\Rightarrow 3(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq(a\sqrt{b}+b\sqrt{c}+c\sqrt{a} )(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}) \geq (a+b+c)^2 \geq 3(ab+bc+ac)[/TEX]
\Rightarrow đpcm
 
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