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\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
\(\Leftrightarrow a+b=a+c+b+c+2\sqrt{\left(a+c\right)\left(b+c\right)}\)
\(\Leftrightarrow2c+2\sqrt{\left(a+c\right)\left(b+c\right)}=0\)
\(\Leftrightarrow c+\sqrt{\left(a+c\right)\left(b+c\right)}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\-c=\sqrt{\left(a+c\right)\left(b+c\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\c^2=\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\ab+bc+ac=0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\left(đúng\right)\)
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
Ta có :
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\left(1+\frac{a+b+c}{a}\right)\left(1+\frac{a+b+c}{b}\right)\left(1+\frac{a+b+c}{c}\right)\)
\(=\left(\frac{2a+b+c}{a}\right)\left(\frac{2b+a+c}{b}\right)\left(\frac{2c+a+b}{c}\right)\)
\(=\left(\frac{a+b}{a}+\frac{a+c}{a}\right)\left(\frac{a+b}{b}+\frac{b+c}{b}\right)\left(\frac{a+c}{c}+\frac{b+c}{c}\right)\)
Áp dụng BĐT Cô-si,ta có :
\(\frac{a+b}{a}+\frac{a+c}{a}\ge2\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{a^2}}\)
\(\frac{a+b}{b}+\frac{b+c}{b}\ge2\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{b^2}}\)
\(\frac{a+c}{c}+\frac{b+c}{c}\ge2\sqrt{\frac{\left(a+c\right)\left(b+c\right)}{c^2}}\)
\(\Rightarrow\left(\frac{a+b}{a}+\frac{a+c}{a}\right)\left(\frac{a+b}{b}+\frac{b+c}{b}\right)\left(\frac{a+c}{c}+\frac{b+c}{c}\right)\ge8\sqrt{\frac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{a^2b^2c^2}}\)
\(\ge8\sqrt{\frac{\left[8\sqrt{a^2b^2c^2}\right]^2}{a^2b^2c^2}}=8\sqrt{64}=64\)
Dấu "=" xảy ra khi a = b = c = \(\frac{1}{3}\)
Áp dụng AM-GM
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3.\dfrac{1}{\sqrt[3]{abc}}=9\)
\(\rightarrow1.\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
vậy ta có điều phải chứng minh
Dấu "=" \(a=b=c=\dfrac{1}{3}\)
Áp dụng svac-xơ:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=9\)
Dấu = xảy ra <=> \(a=b=c=\dfrac{1}{3}\)
C2: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)
\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\)
\(\ge3+2+2+2=9\) (theo cosi)
Dấu = xảy ra <=>\(a=b=c=\dfrac{1}{3}\)
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
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
{1+a}{1+b}+{1+b}{1+c}+{1+c}{1+a}
=1+a+b+ab+1+b+c+bc +1+c+a+ca
=1+1+1+{a+b+c}+{a+b+c} +ab+bc+ca
=5+ab+bc+ca
vìab+bc+ca >0 =>5+ab+bc+ca >5
lik-e cho minh nha