100. 设$a+b+c=1$均为正数,且$ab+bc+ca\le \dfrac{1}{3}$,证明:
（1）$\dfrac{{{a}^{2}}}{b}+\dfrac{{{b}^{2}}}{c}+\dfrac{{{c}^{2}}}{a}\ge 1$
（2）${{a}^{2}}+{{b}^{2}}\ge 2ab$
证明：（1）∵${{a}^{2}}+{{b}^{2}}\ge 2ab$,
${{b}^{2}}+{{c}^{2}}\ge 2bc$,
${{a}^{2}}+{{c}^{2}}\ge 2ac$,
∴${{a}^{2}}+{{b}^{2}}+{{c}^{2}}\ge ab+bc+ca$.
∵${{\left( a+b+c \right)}^{2}}=1$,
即${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca=1$,
∴$3\left( ab+bc+ca \right)\le 1$,即${{a}^{2}}+{{b}^{2}}+{{c}^{2}}\le \dfrac{1}{3}$.
（2） ∵$\dfrac{{{a}^{2}}}{b}+b\ge 2a$,$\dfrac{{{b}^{2}}}{c}+c\ge 2b$,$\dfrac{{{c}^{2}}}{a}+a\ge 2c$,
∴$\dfrac{{{a}^{2}}}{b}+\dfrac{{{b}^{2}}}{c}+\dfrac{{{c}^{2}}}{a}+\left( a+b+c \right)\ge 2\left( a+b+c \right)$,
∴$\dfrac{{{a}^{2}}}{b}+\dfrac{{{b}^{2}}}{c}+\dfrac{{{c}^{2}}}{a}\ge a+b+c=1$.
另证：(Ⅰ)由柯西不等式
$\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{1}^{2}}+{{1}^{2}}+{{1}^{2}} \right)\ge {{\left( a\cdot 1+b\cdot 1+c\cdot 1 \right)}^{2}}$，
即${{a}^{2}}+{{b}^{2}}+{{c}^{2}}\ge \dfrac{1}{3}$，
又${{a}^{2}}+{{b}^{2}}+{{c}^{2}}={{\left( a+b+c \right)}^{2}}-2\left( ab+bc+ca \right)$，
∴$1-2\left( ab+bc+ca \right)\ge \dfrac{1}{3}$，即$ab+bc+ca\le \dfrac{1}{3}$。
(Ⅱ)由柯西不等式
$\begin{align} & \left( a+b+c \right)\left( \dfrac{{{a}^{2}}}{b}+\dfrac{{{b}^{2}}}{c}+\dfrac{{{c}^{2}}}{a} \right) \\ & \ge {{\left( \sqrt{a}\cdot \dfrac{c}{\sqrt{a}}+\sqrt{b}\cdot \dfrac{a}{\sqrt{b}}+\sqrt{c}\cdot \dfrac{b}{\sqrt{c}} \right)}^{2}} \\ \end{align}$
$={{\left( a+b+c \right)}^{2}}=1$
∴$\dfrac{{{a}^{2}}}{b}+\dfrac{{{b}^{2}}}{c}+\dfrac{{{c}^{2}}}{a}\ge 1$
