Câu 3:Áp dụng BĐT Mincopxki ta có:$M\geq \sqrt{(a+b+c)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}$ $\geq\sqrt{(a+b+c)^2+\dfrac{81}{(a+b+c)^2}}$ $=\sqrt{(a+b+c)^2+\dfrac{1296}{(a+b+c)^2}-\dfrac{1215}{(a+b+c)^2}}$ $\geq\sqrt{2\sqrt{(a+b+c)^2.\dfrac{1296}{(a+b+c)^2}}-\dfrac{1215}{36}}=\dfrac{3\sqrt{17}}{2}$$\min M=\dfrac{3\sqrt{17}}{2} \Leftrightarrow a=b=c=2$
Câu 3:Áp dụng BĐT Mincopxki ta có:$M\geq \sqrt{(a+b+c)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}$ $\geq\sqrt{(a+b+c)^2+\dfrac{81}{(a+b+c)^2}}$ $=\sqrt{(a+b+c)^2+\dfrac{1296}{(a+b+c)^2}-\dfrac{1215}{(a+b+c)^2}}$ $\geq\sqrt{2\sqrt{(a+b+c)^2.\dfrac{1296}{(a+b+c)^2}}-\dfrac{1215}{36}}=\dfrac{3}{\sqrt{17}}{2}$$\min M=\dfrac{3\sqrt{17}}{2} \Leftrightarrow a=b=c=2$
Câu 3:Áp dụng BĐT Mincopxki ta có:$M\geq \sqrt{(a+b+c)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}$ $\geq\sqrt{(a+b+c)^2+\dfrac{81}{(a+b+c)^2}}$ $=\sqrt{(a+b+c)^2+\dfrac{1296}{(a+b+c)^2}-\dfrac{1215}{(a+b+c)^2}}$ $\geq\sqrt{2\sqrt{(a+b+c)^2.\dfrac{1296}{(a+b+c)^2}}-\dfrac{1215}{36}}=\dfrac{3\sqrt{17}}{2}$$\min M=\dfrac{3\sqrt{17}}{2} \Leftrightarrow a=b=c=2$