\(P=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\left(\dfrac{a}{a}+\dfrac{b}{b}+\dfrac{c}{c}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
\(\left\{{}\begin{matrix}\dfrac{a}{b}=m\\\dfrac{b}{c}=n\\\dfrac{c}{a}=p\end{matrix}\right.\)
\(P=3+\left(m+\dfrac{1}{m}\right)+\left(n+\dfrac{1}{n}\right)+\left(p+\dfrac{1}{p}\right)\)
c/m: \(Q=k+\dfrac{1}{k}=\left(\left(\sqrt{k}\right)^2-2\sqrt{k}.\dfrac{1}{\sqrt{k}}k+\dfrac{1}{\sqrt{k}}\right)+2=\left(\sqrt{k}-\dfrac{1}{\sqrt{k}}\right)^2+2\ge2\forall k>0\)
\(P\ge3+6=9\) đẳng thức khi a=b=c=1
Do a, b, c là các số dư
=>a, b, c >0
=>a+b+c>0 và \(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)>0
Áp dụng bất đẳng thức Côsi ta có :
a+b+c \(\ge\)3\(\sqrt[3]{abc}\)
\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)\(\ge\)3\(\sqrt[3]{\dfrac{1}{abc}}\)
=> (a+b+c)(\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\))\(\ge\)3\(\sqrt[3]{abc}\).3\(\sqrt[3]{\dfrac{1}{abc}}\)
=> (a+b+c)(\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\))\(\ge\)9\(\sqrt[3]{abc\dfrac{1}{abc}}\)
=> (a+b+c)(\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\))\(\ge\)9
Hay P\(\ge\)9
Dấu "=" xảy ra <=> a=b=c
. Mình cảm ơn trước !
