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Giải:
Áp dụng BĐT Cauchy cho nhiều số dương:
\(1+\dfrac{1}{a}=\dfrac{a+1}{a}=\dfrac{a+a+b+c}{a}\ge\dfrac{4\sqrt[4]{a^2.b.c}}{a}\)
\(1+\dfrac{1}{b}=\dfrac{b+1}{b}=\dfrac{a+b+b+c}{b}\ge\dfrac{4\sqrt[4]{a.b^2.c}}{a}\)
\(1+\dfrac{1}{c}=\dfrac{c+1}{c}=\dfrac{a+b+c+c}{b}\ge\dfrac{4\sqrt[4]{a.b.c^2}}{c}\)
Nhân vế theo vế, được:
\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge\dfrac{64\sqrt[4]{a^4.b^4.c^4}}{a.b.c}\)
\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge\dfrac{64.abc}{abc}\)
\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
Vậy ...
Giải:
\(\dfrac{a}{\left(a+1\right)\left(b+1\right)}+\dfrac{b}{\left(b+1\right)\left(c+1\right)}+\dfrac{c}{\left(c+1\right)\left(a+1\right)}\ge\dfrac{3}{4}\)(*)
\(\Leftrightarrow\) \(\dfrac{a\left(c+1\right)+b\left(a+1\right)+c\left(b+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\) \(\dfrac{ac+a+ab+b+bc+c}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\) \(\ge\) \(\dfrac{3}{4}\)
Do a+1 ; b+1; c+1 >0
\(\Rightarrow\) 4ac+4a+4ab+4b+4bc+4c \(\ge\) 3abc+3ac+3bc+3ab+3a+3b+3c+3
\(\Leftrightarrow\) ac+ab+bc+a+b+c -6 \(\ge\) 0
Áp dụng BĐT Cô-si cho 3 số
Ta có: a+b+c \(\ge\) \(3\sqrt[3]{abc}=3\)
ab+bc+ca \(\ge\) \(3\sqrt[3]{\left(abc\right)^2}\) = 3
\(\Rightarrow\)ac+ab+bc+a+b+c -6 \(\ge\) 0 ( luôn đúng)
\(\Rightarrow\) (*) được chứng minh
Dấu "=" xảy ra \(\Leftrightarrow\) a=b=c=1
Viết gọn lại, ta cần chứng minh:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}\right)\)
\(\Leftrightarrow\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{a+b}{ab}}\right)=\sum\dfrac{4ab}{a+b}\)
Thật vậy, ta có:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum\left(2\sqrt{\left(a+b\right).\dfrac{1}{4}}\right)^2=\sum a+b\)
Vậy ta cần chứng minh:
\(\sum a+b\ge\sum\dfrac{4ab}{a+b}\Leftrightarrow\sum\left(a+b\right)^2\ge\sum4ab\Leftrightarrow\sum\left(a-b\right)^2\ge0\)
Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c
Lời giải:
Ta có:
\(\text{VT}=\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\)
\(=\frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}=\frac{ab+bc+ac+a+b+c}{abc+(ab+bc+ac)+(a+b+c)+1}\)
\(=\frac{ab+bc+ac+a+b+c}{2+(a+b+c)+ab+bc+ac}\)
Ta cần chứng minh \(\text{VT}\geq \frac{3}{4}\)
\(\Leftrightarrow \frac{ab+bc+ac+a+b+c}{2+(a+b+c)+ab+bc+ac}\geq \frac{3}{4}\)
\(\Leftrightarrow 4(ab+bc+ac+a+b+c)\geq 3(ab+bc+ac+a+b+c)+6\)
\(\Leftrightarrow ab+bc+ac+a+b+c\geq 6\)
\(\Leftrightarrow ab+bc+ac+a+b+c\geq 6\sqrt[6]{ab.bc.ac.a.b.c}\)
(Đúng theo BĐT Cô-si)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=1\)
\(VT=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)=\left(1+\dfrac{1}{b}+\dfrac{1}{a}+\dfrac{1}{ab}\right)\left(1+\dfrac{1}{c}\right)=1+\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{bc}+\dfrac{1}{a}+\dfrac{1}{ac}+\dfrac{1}{ab}+\dfrac{1}{abc}=1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{abc}\) Áp dụng BĐT Cauchy nhiều lần , ta có :
\(a+b+c\ge3\sqrt[3]{abc}\Leftrightarrow\left(\dfrac{a+b+c}{3}\right)^3\ge abc\Leftrightarrow\dfrac{1}{27}\ge abc\Leftrightarrow\dfrac{1}{abc}\ge27\)
\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge3\sqrt[3]{\dfrac{1}{ab}.\dfrac{1}{bc}.\dfrac{1}{ac}}=3\sqrt[3]{\left(\dfrac{1}{abc}\right)^2}\ge3\sqrt[3]{27.27}=27\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\ge3\sqrt[3]{27}=9\)
\(\Rightarrow VT\ge27+9+27+1=64\)
\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
\("="\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Ta có:
\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)=\left(1+\dfrac{a+b+c}{a}\right)\left(1+\dfrac{a+b+c}{b}\right)\left(1+\dfrac{a+b+c}{c}\right)\)
\(=\left(\dfrac{2a+b+c}{a}\right)\left(\dfrac{a+2b+c}{b}\right)\left(\dfrac{a+b+2c}{c}\right)\)
\(=\left(\dfrac{a+b}{a}+\dfrac{a+c}{a}\right)\left(\dfrac{a+b}{b}+\dfrac{b+c}{b}\right)\left(\dfrac{a+c}{c}+\dfrac{b+c}{c}\right)\)
Áp dụng bất đẳng thức Cô - si ta có:
\(\dfrac{a+b}{a}+\dfrac{a+c}{a}\ge2\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{a^2}}\)
\(\dfrac{a+b}{b}+\dfrac{b+c}{b}\ge2\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)}{b^2}}\)
\(\dfrac{a+c}{c}+\dfrac{b+c}{c}\ge2\sqrt{\dfrac{\left(a+c\right)\left(b+c\right)}{c^2}}\)
\(\Rightarrow\left(\dfrac{a+b}{a}+\dfrac{a+c}{a}\right)\left(\dfrac{a+b}{b}+\dfrac{b+c}{b}\right)\left(\dfrac{a+c}{c}+\dfrac{b+c}{c}\right)\ge8\sqrt{\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2}{a^2b^2c^2}}=8.\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Áp dụng bất đẳng thức Cô - si ta có:
\(a+b\ge2\sqrt{ab}\)
\(b+c\ge2\sqrt{bc}\)
\(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{a^2b^2c^2}=8abc\)
\(8.\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\ge8\sqrt{a^2b^2c^2}=8.\dfrac{8abc}{abc}=64\)
hay \(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\) (đpcm)
1) Áp dụng BĐT Cô si
ta có
\(\left(\sqrt{a+b}-\dfrac{1}{2}\right)^2\ge0\forall a,b\inĐK\)
\(\Leftrightarrow a+b-2\sqrt{a+b}.\dfrac{1}{2}+\dfrac{1}{4}\ge0\)
\(\Leftrightarrow a+b+\dfrac{1}{4}\ge\sqrt{a+b}\)
vậy ĐPCM
Bài 2
Áp dụng bđt Cauchy ta có \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\Rightarrow\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}\le\dfrac{\sqrt{ab}}{2}\)
Thiết lập tương tự và thu lại ta có:
\(\Rightarrow VP\le4\left(\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2}\right)=2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(1\right)\)
Áp dụng bđt Cauchy ta có \(a+b\ge2\sqrt{ab}\)
\(\Rightarrow\left(a+b+\dfrac{1}{2}\right)^2\ge\left(2\sqrt{ab}+\dfrac{1}{2}\right)^2\ge2.2\sqrt{ab}.\dfrac{1}{2}=2\sqrt{ab}\)
Thiết lập tương tự và thu lại ta có:
\(\Rightarrow VT\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow VT\ge VP\)
\(\Rightarrowđpcm\)
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
Lời giải:
Áp dụng BĐT AM-GM:
\(\frac{a^4}{(a+2)(b+2)}+\frac{a+2}{27}+\frac{b+2}{27}+\frac{1}{9}\geq 4\sqrt[4]{\frac{a^4}{27.27.9}}=\frac{4a}{9}\)
\(\frac{b^4}{(b+2)(c+2)}+\frac{b+2}{27}+\frac{c+2}{27}+\frac{1}{9}\geq \frac{4b}{9}\)
\(\frac{c^4}{(c+2)(a+2)}+\frac{c+2}{27}+\frac{a+2}{27}+\frac{1}{9}\geq \frac{4c}{9}\)
Cộng theo vế và rút gọn:
\(\frac{a^4}{(a+2)(b+2)}+\frac{b^4}{(b+2)(c+2)}+\frac{c^4}{(c+2)(a+2)}+\frac{2(a+b+c)}{27}+\frac{7}{9}\geq\frac{4(a+b+c)}{9}\)
\(\frac{a^4}{(a+2)(b+2)}+\frac{b^4}{(b+2)(c+2)}+\frac{c^4}{(c+2)(a+2)}\geq \frac{10(a+b+c)}{27}-\frac{7}{9}=\frac{30}{27}-\frac{7}{9}=\frac{1}{3}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
\(a+b+c=1=>\left\{{}\begin{matrix}1-a=b+c\\1-b=a+c\\1-c=a+b\\\end{matrix}\right.\)
\(=>A=\left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right)=\left(\dfrac{1-a}{a}\right)\left(\dfrac{1-b}{b}\right)\left(\dfrac{1-c}{c}\right)\)
\(=\left(\dfrac{b+c}{a}\right)\left(\dfrac{a+c}{b}\right)\left(\dfrac{a+b}{c}\right)\)
bbđt AM-GM
\(=>A\ge\dfrac{2\sqrt{bc}.2\sqrt{ac}.2\sqrt{ab}}{abc}=\dfrac{8abc}{abc}=8\left(đpcm\right)\)
dấu"=" xảy ra<=>\(a=b=c=\dfrac{1}{3}\)
Đặt vế trái BĐT cần chứng minh là P
Ta có:
\(P=\left(\dfrac{a+b+c}{a}-1\right)\left(\dfrac{a+b+c}{b}-1\right)\left(\dfrac{a+b+c}{c}-1\right)\)
\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\ge\dfrac{2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}}{abc}=8\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)