Question

Cho x,y, z la cac so duong thoa man dieu kien x+y+z=a

tim GTNN : Q=\(\left(1+\dfrac{a}{x}\right)\left(1+\dfrac{a}{y}\right)\left(1+\dfrac{a}{z}\right)\)

Answer 1

Q=\(\left(1+\dfrac{a}{x}\right)\left(1+\dfrac{a}{y}\right)\left(1+\dfrac{a}{z}\right)\)

\(Q=\left(\dfrac{x+a}{x}\right)\left(\dfrac{y+a}{y}\right)\left(\dfrac{z+a}{z}\right)\)\

=\(\left(\dfrac{2x+y+z}{x}\right)\left(\dfrac{2y+x+z}{y}\right)\left(\dfrac{2z+x+y}{z}\right)\)

=\(\dfrac{\left(2x+y+z\right)\left(2y+x+z\right)\left(2z+x+y\right)}{xyz}\)

ÁP dụng BĐT cô si

\(2x+y+z=x+x+y+z\ge4\sqrt[4]{x^2yz}\)

\(2y+x+z=y+y+x+z\ge4\sqrt[4]{y^2xy}\)

\(2z+y+x=z+z+x+y\ge4\sqrt[4]{z^2xy}\)

=> Q\(\ge\dfrac{64.\sqrt[4]{x^4y^4z^4}}{xyz}=64\)

=> MinQ=64 khi x=y=z=a/3