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Áp dụng bđt Cauchy - Schwarz dạng engel (full name nhé) , ta có
\(B=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{\left(1+1+1\right)^2}{1+a+1+b+1+c}=\frac{9}{3+a+b+c}\ge\frac{9}{3+3}=\frac{3}{2}\)
Đẳng thức xảy ra <=> \(a=b=c=1\)
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)
\(\Rightarrow3.P\ge9\Rightarrow P\ge3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng bđt Cauchy , ta có :
\(P=\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge2\sqrt{a}.2\sqrt{b}.2\sqrt{c}=8\sqrt{abc}=8\)
Dấu "=" xảy ra khi a = b = c = 1
Vậy Min P = 8 <=> a = b = c = 1
b) Vì \(\left|a\right|=\left|-a\right|\)\(\Rightarrow\)\(\left|x-2020\right|=\left|2020-x\right|\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)biểu thức P(x), ta có:
\(\left|2020-x\right|+\left|x+2021\right|\ge\left|2020-x+x+2021\right|=4041\)
\(\Rightarrow\)\(P\left(x\right)\ge4041\)
Dấu "=" xảy ra khi và chỉ khi: \(\left(2020-x\right)\left(x+2021\right)>0\)
\(\Leftrightarrow-2021< x< 2020\)
Vậy \(P\left(x\right)_{min}=4041\)\(\Leftrightarrow\)\(-2021< x< 2020\)
\(B=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\)
\(\ge3\sqrt[3]{\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Dễ có:\(\left(1+a\right)\left(1+b\right)\left(1+c\right)\le\left(\frac{3+a+b+c}{3}\right)^3\le8\)
Khi đó \(B\ge\frac{3}{2}\)
Đẳng thức xảy ra tại a=b=c=1
a) ĐKXĐ: \(x\ne1\)
b) \(A=\frac{2}{x-1}+\frac{2\left(x+1\right)}{x^2+x+1}+\frac{x^2-10x+3}{x^3-1}\)
\(=\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-10x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-10x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{5x^2-8x+3}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(x-1\right)\left(5x-3\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{5x-3}{x^2+x+1}\)
Vì a;b;c > 0 nên \(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}>0\)
BĐT Cosi :
\(9a+\dfrac{1}{a}\ge2.\sqrt{9a.\dfrac{1}{a}}=2.3=6\\ 9b+\dfrac{1}{b}\ge6\\ 9c+\dfrac{1}{c}\ge6\\ \Rightarrow\left(9a+\dfrac{1}{a}\right)+\left(9b+\dfrac{1}{b}\right)+\left(9c+\dfrac{1}{c}\right)\ge18\\ \Rightarrow9\left(a+b+c\right)+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge18\\ \Rightarrow9+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge18\\ \Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
Dấu "=" xảy ra khi a=b=c=1/3
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{1}=9\)
Áp dụng BĐT Cauchy ta có:
\(a+1\ge2\sqrt{a.1}=2\sqrt{a}\)
\(b+1\ge2\sqrt{b.1}=2\sqrt{b}\)
\(c+1\ge2\sqrt{c.1}=2\sqrt{c}\)
Dấu "=" xảy ra <=> \(a=b=c=1\)
\(P=\left(a+1\right)\left(b+1\right)\left(c+1\right)\) \(\ge\)\(2\sqrt{a}.2\sqrt{b}.2\sqrt{c}=8.\sqrt{abc}=8\)
Vậy Min P = 8 <=> a = b = c = 1
Cauchy :
\(P=\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge2\sqrt{a}.2\sqrt{b}.2\sqrt{c}=8.\sqrt{abc}=8\)
Đẳng thức xảy ra <=> a = b = c = 1