$A=(\frac{a^2}{b^2}+\frac{b^2}{a^2}+2)-3(\frac{a}{b}+\frac{b}{a})+2=(\frac{a}{b}+\frac{b}{a}-1)(\frac{a}{b}+\frac{b}{a}-2)$ Ta có: $\frac{x}{y}+\frac{y}{x}\geq 2\forall x,y> 0\Rightarrow \frac{a}{b}+\frac{b}{a}-1\geq 1; \frac{a}{b}+\frac{b}{a}-2\geq 0$$\Rightarrow A\geq 0 (đpcm)$
$A=(\frac{a^2}{b^2}+\frac{b^2}{a^2}+2)-3(\frac{a}{b}+\frac{b}{a})+2=(\frac{a}{b}+\frac{b}{a}-1)(\frac{a}{b}+\frac{b}{a}-2)$ Ta có: $\frac{x}{y}+\frac{y}{x}\geq 2\forall x,y> 0\Rightarrow \frac{a}{b}+\frac{b}{a}-1>0; \frac{a}{b}+\frac{b}{a}-2\geq 0$$\Rightarrow A\geq 0 (đpcm)$
$A=(\frac{a^2}{b^2}+\frac{b^2}{a^2}+2)-3(\frac{a}{b}+\frac{b}{a})+2=(\frac{a}{b}+\frac{b}{a}-1)(\frac{a}{b}+\frac{b}{a}-2)$ Ta có: $\frac{x}{y}+\frac{y}{x}\geq 2\forall x,y> 0\Rightarrow \frac{a}{b}+\frac{b}{a}-1
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eq 1; \frac{a}{b}+\frac{b}{a}-2\geq 0$$\Rightarrow A\geq 0 (đpcm)$