代数恒等变形

egin{example}
(三元均值不等式)设$a,b,cin mathbb{R}^+$,则
$$
frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}le sqrt[3]{abc}le frac{a+b+c}{3}le sqrt{frac{a^2+b^2+c^2}{3}},
$$
end{example}
egin{solution}
先证明:
$$
a^3+b^3+c^3ge 3abc,quad a,b,cin mathbb{R}^+.
$$
事实上,
egin{align*}
a^3+b^3+c^3-3abc &=left( a+b ight) ^3+c^3-3a^2b-3ab^2-3abc
\
&=left( a+b+c ight) left[ left( a+b ight) ^2-left( a+b ight) c+c^2 ight] -3ableft( a+b+c ight)
\
&=left( a+b+c ight) left[ left( a+b ight) ^2-left( a+b ight) c+c^2-3ab ight]
\
&=left( a+b+c ight) left( a^2+b^2+c^2-ab-bc-ca ight)
\
&=left( a+b+c ight) left[ frac{1}{2}left( a-b ight) ^2+frac{1}{2}left( b-c ight) ^2+frac{1}{2}left( c-a ight) ^2 ight] ge 0.
end{align*}
令$a^3 o a,b^3 o b,c^3 o c$即可.
end{solution}

egin{example}
(2012年初联)已知互不相等的实数$a,b,c$满足$a+frac{1}{b}=b+frac{1}{c}=c+frac{1}{a}=t$,求$t$.
end{example}
egin{solution}
由$a+frac{1} {b}=t$得$b=frac{1}{t-a}$,代入$b+frac{1}{c}=t$得$frac{1}{t-a}+frac{1}{c}=t$,整理得$ct^2-(ac+1)t+(a-c)=0$ding{172}.

又由$c+frac{1}{a}=t$可得$ac+1=at$,代入ding{172}式得$ct^2-at^2+(a-c)=0$,即$(c-a)(t^2-1)=0$,又$c eq a$,所以$t^2-1=0$,所以$t=pm 1$.

验证可知: $b=frac{1}{1-a},c=frac{a-1}{a}$时$t=1$;
$b=-frac{1}{1+a},c=-frac{a+1}{a}$时$t=-1$.因此, $t=pm 1$.
end{solution}