Cho x,y,z>0 thoả mãn x+y+z=<1. Tìm min của 2(x+y+z)+3(1/x+1/y+1/z)
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18 tháng 4 2020
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24 tháng 7 2019
Ta có \(1+\frac{a}{x}=1+\frac{x+y+z}{x}=\frac{2x+y+z}{x}\)
Áp dụng BĐT cosi \(x+x+y+z\ge4\sqrt[4]{x^2yz}\)
=> \(1+\frac{a}{x}\ge\frac{4\sqrt[4]{x^2yz}}{x}\)
Tương tự\(1+\frac{a}{y}\ge\frac{4\sqrt[4]{y^2xz}}{y}\); \(1+\frac{a}{z}\ge\frac{4\sqrt[4]{z^2yx}}{z}\)
=> \(Q\ge\frac{64.\sqrt[4]{x^4y^4z^4}}{xyz}=64\)
MinQ=64 khi \(x=y=z=\frac{a}{3}\)