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Áp dụng bđt cô si cho 2 số dương ta có:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc\cdot\frac{1}{abc}}=9\)
Cách 1. Áp dụng bđt Bunhiacopxki : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(\sqrt{a.\frac{1}{a}}+\sqrt{b.\frac{1}{b}}+\sqrt{c.\frac{1}{c}}\right)^2=\left(1+1+1\right)^2=9\)
Cách 2. Áp dụng bđt Cauchy :
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Áp dụng bất đẳng thức Cô-si cho 3 số ta được
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
Nhân 2 vế của bất đẳng thức trên lại ta được đpcm
Dấu ''='' <=> a = b = c
ko dùng đến BĐT cauchy cx dc!
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)
\(=1+1+1+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}\)
\(=3+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)\)
Ta có:\(\frac{a}{c}+\frac{c}{a}\ge2\),thật vậy:
Gỉa sử \(a\ge c\),khi đó:\(a=c+m\)
\(\Rightarrow\frac{a}{c}+\frac{c}{a}=\frac{c+m}{c}+\frac{c}{c+m}=1+\frac{m}{c}+\frac{c}{c+m}\ge1+\frac{m}{c+m}+\frac{c}{c+m}=1+\frac{m+c}{m+c}=1+1=2\)
Chứng minh tương tự,ta được:
\(\hept{\begin{cases}\frac{c}{b}+\frac{b}{c}\ge2\\\frac{a}{b}+\frac{b}{a}\ge2\end{cases}}\)
\(\Rightarrow\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{c}{b}+\frac{b}{c}\ge6\)
\(\Rightarrow3+\frac{a}{b}+\frac{b}{a}+\frac{c}{a}+\frac{a}{c}+\frac{c}{b}+\frac{b}{c}\ge9\left(đpcm\right)\)
Đặt A là biểu thức cần CM
ví dụ Từ ĐK a + b + c = 3 => a² + b² + c² ≥ 3 ( Tự chứng minh )
Áp dụng BĐT quen thuộc x² + y² ≥ 2xy
a^4 + b² ≥ 2a²b (1)
b^4 + c² ≥ 2b²c (2)
c^4 + a² ≥ 2c²a (3)
Đây nhé
Đặt b + c = x ; c + a = y ; a + b = z
\(\Rightarrow\hept{\begin{cases}x+y=2c+b+a=2c+z\\y+z=2a+b+c=2a+x\\x+z=2b+a+c=2b+y\end{cases}}\)
\(\Rightarrow\frac{x+y-z}{2}=c;\frac{y+z-x}{2}=a;\frac{x+z-y}{2}=b\)
Thay vào PT đã cho ở đề bài , ta có :
\(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\)
\(=\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}-3\right)\)
\(\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)
( cái này cô - si cho x/y + /x ; x/z + z/x ; y/z + z/y)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)
\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
Áp dung BĐT cô si cho 2 số không âm ta được:
\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\)
\(\frac{a}{c}+\frac{c}{a}\ge2\sqrt{\frac{a}{c}.\frac{c}{a}}=2\)
\(\frac{b}{c}+\frac{c}{b}\ge2\sqrt{\frac{b}{c}.\frac{c}{b}}=2\)
Suy ra: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3+2+2+2=9\left(\text{ điều phải chứng minh}\right)\)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+b.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+c.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)
\(=\left(1+1+1\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
Áp dụng tổng hai phân số nghịch đảo lớn hơn hoặc bằng 2 ta có :
\(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge3+2+2+2=9\)
=> ĐPCM
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\left(\frac{9}{a+b+c}\right)=9\left(dpcm\right)\)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+\frac{a}{b}+\frac{c}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}.\text{ÁP DỤNG BĐT CÔ SI TA ĐƯỢC:}\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge3+2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{bc}{bc}}+2\sqrt{\frac{c}{a}.\frac{a}{c}}=3+2+2+2=9\)
Lời giải
Ta có: \(\left(a+b+\frac{1}{4}\right)^2=\frac{1}{16}\left(4a+4b-1\right)^2+\left(a+b\right)\ge a+b\)
Tương tự: \(\left(b+c+\frac{1}{4}\right)^2\ge b+c;\left(c+a+\frac{1}{4}\right)^2\ge c+a\)
Như vậy: \(L.H.S\left(VT\right)\ge\left(a+b\right)+\left(b+c\right)+\left(c+a\right)=\left(\frac{1}{\frac{1}{a}}+\frac{1}{\frac{1}{b}}\right)+\left(\frac{1}{\frac{1}{b}}+\frac{1}{\frac{1}{c}}\right)+\left(\frac{1}{\frac{1}{c}}+\frac{1}{\frac{1}{a}}\right)\)
\(\ge4\left(\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}+\frac{1}{\frac{1}{c}+\frac{1}{a}}\right)=R.H.S\left(VP\right)\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{8}\). Ta có đpcm.
khác cách tth xíu
Ta có:
\(VP=\Sigma_{cyc}\frac{4}{\frac{1}{a}+\frac{1}{b}}\le\Sigma_{cyc}\frac{4}{\frac{4}{a+b}}=2\left(a+b+c\right)\)
Gio ta di chung minh
\(VT\ge2\left(a+b+c\right)\)
Ta lai co:
\(VT=\Sigma_{cyc}\left(a+b+\frac{1}{4}\right)^2\ge\frac{\left[2\left(a+b+c\right)+\frac{3}{4}\right]^2}{3}\)
Chung minh
\(\frac{\left[2\left(a+b+c\right)+\frac{3}{4}\right]^2}{3}\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\left[2\left(a+b+c\right)-\frac{3}{4}\right]^2\ge0\) (đúng)
Dau '=' xay ra khi \(a=b=c=\frac{1}{8}\)
Ta có :
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}\)
\(=1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)
\(=3+\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\right)\)
\(\frac{1}{6}\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\right)\ge\sqrt[6]{\frac{a}{b}.\frac{a}{c}.\frac{b}{a}.\frac{b}{c}.\frac{c}{a}.\frac{c}{b}}\)
\(\Rightarrow\frac{1}{6}\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\right)\ge\sqrt[6]{1}\)
\(\Rightarrow\frac{1}{6}\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\right)\ge1\)
\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\ge1:\frac{1}{6}=6\)
\(\Rightarrow3+\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\right)\ge3+6=9\)
~
Còn 1 cách dùng BĐT Cauchy:
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\)
\(=3+\left[\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\right]\)
Áp dụng BĐT Cauchy cho \(\frac{a}{b}+\frac{b}{a};\frac{a}{c}+\frac{c}{a};\frac{b}{c}+\frac{c}{b};\)có :
\(\left(\frac{a}{b}+\frac{b}{a}\right)+\ge2\)
\(\left(\frac{b}{c}+\frac{c}{b}\right)\ge2\)
\(\left(\frac{a}{c}+\frac{c}{a}\right)\ge2\)
\(\Rightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\ge2+2+2=6\)
Tương tự, bạn làm tiếp.