1.ta có a+√ab+3√abc=a+√12a.2b+3√14a.b.4c ≤a+14a+b+112a+b3+43c=43(a+b+c)⇒P≥32(a+b+c)−3√a+b+c Đặt t=√a+b+c,t>0P ≥32t2−3t=32(1t−1)2−32≥−32Dấu '=' $\Leftrightarrow a=\frac{16}{21};b=\frac{4}{21};c=\frac{1}{21}4
1.ta có
a+√ab+3√abc=a+√12a.2b+3√14a.b.4c ≤a+14a+b+112a+b3+43c=43(a+b+c)⇒P≥32(a+b+c)−3√a+b+c Đặt
t=√a+b+c,t>0P
≥32t2−3t=32(1t−1)2−32≥−32Dấu '=' $\Leftrightarrow a=\frac{16}{21};b=\frac{4}{21};c=\frac{1}{21}
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