A=(a2b2+b2a2+2)−3(ab+ba)+2=(ab+ba−1)(ab+ba−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$$\Rightarrow A\geq 0 (đpcm)$
A=(a2b2+b2a2+2)−3(ab+ba)+2=(ab+ba−1)(ab+ba−2) Ta có: $\frac{x}{y}+\frac{y}{x}\geq 2\forall x,y> 0\Rightarrow \frac{a}{b}+\frac{b}{a}-1
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eq 1; \frac{a}{b}+\frac{b}{a}-2\geq 0$$\Rightarrow A\geq 0 (đpcm)$