Câu Tổng hợp đề báo toán số 3

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kimxakiem2507

[TEX]t=x+y+x =3xyz\le \frac{\(x+y+z\)^3}{9}\righ t\ge 3[/TEX]

[TEX]A\ge \frac{t}{3}+\frac{3}{t^2}[/TEX]

[TEX]A\ge \frac{t}{9}+\frac{t}{9}+\frac{3}{t^2}+\frac{t}{9} \ge 1+\frac{1}{3}=\frac{4}{3}[/TEX]
 
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