[Toán 9]Bất đẳng thức

H

harrypham

Đặt
[TEX]B = \frac{b^4}{(a+b)(a^2+b^2)}+ \frac{c^4}{(b+c)(b^2+c^2)}+ \frac{a^4}{(c+a)(c^2+a^2)} [/TEX]​
Ta có

[TEX]A - B = \frac{a^4-b^4}{(a+b)(a^2+b^2)}+ \frac{b^4-c^4}{(b+c)(b^2+c^2)}+ \frac{c^4-a^4}{(c+a)(c^2+a^2)}[/TEX]

[TEX]= a-b+b-c+c-a = 0[/TEX]

[TEX]\Rightarrow A=B[/TEX]​

Vậy

[TEX]2A = \frac{a^4+b^4}{(a+b)(a^2+b^2)}+ \frac{b^4+c^4}{(b+c)(b^2+c^2)}+ \frac{c^4+a^4}{(c+a)(c^2+a^2)} \ge \frac{a+b}{4}+\frac{b+c}{4}+\frac{c+a}{4} = \frac{a+b+c}{2} = 2[/TEX]

Do đó [TEX]\fbox{A \ge 1}[/TEX].
 
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