Đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=1\)
\(P=\sum\dfrac{a}{\sqrt{1+a^2}}=\sum\dfrac{a}{\sqrt{ab+bc+ca+a^2}}=\sum\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\sum\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)=\dfrac{3}{2}\)
\(P_{max}=\dfrac{3}{2}\) khi \(x=y=z=\sqrt{3}\)
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