Áp dụng BĐT Cauchy cho 2 số không âm, ta có:
\(\dfrac{a^2}{b}+b\ge2\sqrt{\dfrac{a^2}{b}.b}=2a\)
\(\dfrac{b^2}{c}+c\ge2\sqrt{\dfrac{b^2}{c}.c}=2b\)
\(\dfrac{c^2}{a}+a\ge2\sqrt{\dfrac{c^2}{a}.a}=2c\)
Cộng từng vế BĐT ta được:
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+b+a+c\ge2a+2b+2c\)
\(\Leftrightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)
=> ĐPCM
:D bác @DƯƠNG PHAN KHÁNH DƯƠNG ủng hộ bạn chủ tus cách khác thì cháu cx góp vui ak :v
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)
\(\Rightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+a+b+c-2a-2b-2c\ge0\)
\(\Rightarrow\left(\dfrac{a^2}{b}-2a+b\right)+\left(\dfrac{b^2}{c}-2b+c\right)+\left(\dfrac{c^2}{a}-2c+a\right)\ge0\)
\(\Rightarrow\left(\dfrac{a}{\sqrt{b}}-\sqrt{b}\right)^2+\left(\dfrac{b}{\sqrt{c}}-\sqrt{c}\right)^2+\left(\dfrac{c}{\sqrt{a}}-\sqrt{a}\right)^2\ge0\) (đúng)
\("="\Leftrightarrow a=b=c\)
Cách khác :V
Theo BĐT Cauchy schwar ta có :
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)