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LV
1
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 6 2021
Từ giả thiết: \(\left\{{}\begin{matrix}a+b+c\ge3\sqrt[3]{abc}=3\\ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\end{matrix}\right.\) (1)
Ta có:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\ge3\left(a+b+c\right)-1\)
Nên ta chỉ cần chứng minh:
\(3\left(a+b+c\right)-1\ge2\left(1+a+b+c\right)\)
\(\Leftrightarrow a+b+c\ge3\) (hiển nhiên đúng theo (1))
12 tháng 9 2021
\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
\(\Leftrightarrow a+b=a+c+b+c+2\sqrt{\left(a+c\right)\left(b+c\right)}\)
\(\Leftrightarrow2c+2\sqrt{\left(a+c\right)\left(b+c\right)}=0\)
\(\Leftrightarrow c+\sqrt{\left(a+c\right)\left(b+c\right)}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\-c=\sqrt{\left(a+c\right)\left(b+c\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\c^2=\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\ab+bc+ac=0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\left(đúng\right)\)
12 tháng 9 2021
Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0
Khi đó:
(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)
=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2
=a+b+2c+2|c|=a+b+2c+2|c|
Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c
Do đó:
(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b
⇒√a+c+√b+c=√a+b
KN
0
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1
Ta có:
\(\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\left(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\)
\(=\dfrac{c}{a-b}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)+\dfrac{a}{b-c}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)+\dfrac{b}{c-a}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\)
Xét:
\(\dfrac{c}{a-b}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\)
\(=1+\dfrac{c}{a-b}\left[\dfrac{b\left(b-c\right)+a\left(c-a\right)}{ab}\right]=1+\dfrac{c}{a-b}\left(\dfrac{b^2-bc+ac-a^2}{ab}\right)\)
\(=1+\dfrac{c}{a-b}\left[\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab}\right]=1+\dfrac{c}{a-b}.\dfrac{\left(b-a\right)\left(a+b-c\right)}{ab}\)
\(=1-\dfrac{c\left(a+b-c\right)}{ab}=1-\dfrac{c.\left(-2c\right)}{ab}=1+\dfrac{2c^2}{ab}\) (do \(a+b+c=0\Rightarrow a+b=-c\))
Tương tự:
\(\dfrac{a}{b-c}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{2a^2}{bc}\)
\(\dfrac{b}{c-a}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{2b^2}{ca}\)
\(\Rightarrow P=3+2\left(\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\right)=3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}\)
Mặt khác ta có đằng thức quen thuộc:
Khi \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)
\(\Rightarrow P=3+\dfrac{2.3abc}{abc}=9\)
PL
10 tháng 1 2021
Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0
Khi đó:
(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)
=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2
=a+b+2c+2|c|=a+b+2c+2|c|
Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c
Do đó:
(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b
⇒√a+c+√b+c=√a+b
PL
10 tháng 1 2021
Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0
Khi đó:
(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)
=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2
=a+b+2c+2|c|=a+b+2c+2|c|
Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c
Do đó:
(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b
⇒√a+c+√b+c=√a+b