Áp dụng bđt bunhiacopski ta có
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\ge\left(ax+by+cz\right)^2\Rightarrow\left(a+b+c\right)^2=\left(\frac{a\sqrt{x}}{\sqrt{x}}+\frac{b\sqrt{y}}{\sqrt{y}}+\frac{c\sqrt{z}}{\sqrt{z}}\right)^2\le\left[\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2+\left(\frac{c}{\sqrt{z}}\right)^2\right]\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]=\left(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\right)\left(x+y+z\right)\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)(Dấu '=' xảy ra khi \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Áp dụng bđt trên ta có \(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c=1\Leftrightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge1\)Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\\a+b+c=1\end{matrix}\right.\)\(\Leftrightarrow a=b=c=\frac{1}{3}\)
Vậy \(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge1\)
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
