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Find the greatest value of xyz for positivevalues of x, y, z subject to the conditionxy + yz + zx = 12.

hjh , 8 Years ago
Grade 12
anser 2 Answers
hjh

Last Activity: 8 Years ago

as the val

jagdish singh singh

Last Activity: 8 Years ago

\hspace{-0.74 cm}$ Given $xy+yz+zx=12\;,x,y,z>0.$ Then Using $\bf{A.M\geq G.M}$\\\\ So $\frac{xy+yz+zx}{3}\geq (xyz)^{\frac{2}{3}}\Rightarrow (xyz)\leq 4^{\frac{3}{2}}=8$\\\\and equality hold when $xy=yz=zx.$

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