Question

cho a,b,c>0 cm \(\dfrac{a+b}{\sqrt{c}}\) +\(\dfrac{b+c}{\sqrt{a}}\) +\(\dfrac{c+a}{\sqrt{b}}\)≥2(\(\sqrt{a}+\sqrt{b}+\sqrt{c}\))

 

Answer 1

đặt \(\left\{{}\begin{matrix}\sqrt{a}=x\\\sqrt{b}=y\\\sqrt{c}=z\end{matrix}\right.\)

=>đề bài đưa về

\(=>P=\dfrac{x^2+y^2}{z}+\dfrac{y^2+z^2}{x}+\dfrac{x^2+z^2}{y}\ge2\left(x+y+z\right)\)

có \(P=\dfrac{x^2}{z}+\dfrac{y^2}{z}+\dfrac{y^2}{x}+\dfrac{z^2}{x}+\dfrac{x^2}{y}+\dfrac{z^2}{y}\ge\dfrac{\left(2x^{ }+2y^{ }+2z^{ }\right)^2}{2\left(x+y+z\right)}\)

\(=\dfrac{4\left(x+y+z\right)^2}{2\left(x+y+z\right)}=2\left(x+y+z\right)\left(dpcm\right)\)

dấu"=" xảy ra<=>x=y=z