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  1. haiproh

    Toán 9 $$a^2b+b^2c+c^2a+abc+\frac{1}{2}abc(3-ab-bc-ca)\leq 4$$

    Cho a,b,c\geq 0: a+b+c=3. Chứng minh rằng: a^2b+b^2c+c^2a+abc+\frac{1}{2}abc(3-ab-bc-ca)\leq 4
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