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ta có (a+b+c ) 2 = a2+b2+c2+2(ab+bc+ac)
Mà a2+b2+c2 >/ ab+bc+ac ( Bạn tự CM: nhân 2 vế với 2 rồi chuyển vế dưa về HDT)
=> (a+b+c ) 2 = 3(ab+bc+ac) => \(a+b+c\ge3\frac{ab+bc+ca}{a+b+c}\)mà a+b+c=abc
\(a+b+c\ge3\frac{ab+bc+ca}{abc}\)
\(a+b+c\ge3.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
2. Áp dụng bất đẳng thức Cô - si cho 3 số dương \(\frac{a}{b},\frac{b}{c},\frac{c}{a}\)ta có
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}\)\(=3\)
Dấu "=" xảy ra <=> a = b = c
cho 1/a+1/b+1/c=2 va :a+b+c=abc
.chung minh rang:
.
\(VT=\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}+\Sigma\frac{a^2}{a^2\left(b+c\right)}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\Sigma a^2\left(b+c\right)+2abc}=\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
cauchy-schwarz:
\(VT=\frac{c^2}{ac^2+bc^2}+\frac{a^2}{a^2b+a^2c}+\frac{b^2}{b^2c+b^2a}+\frac{\sqrt[3]{a^2b^2c^2}}{2abc}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
=> x+y+z=0
Có \(x^3+y^3+z^3-3xyz\)
=\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
=\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2-3xy\right]\)
=0( do x+y+z=0)
=> \(x^3+y^3+z^3=3xyz\)
<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Ta có 1 + ab2 \(\ge\)\(2b\sqrt{a}\)
1 + bc2 \(\ge2c\sqrt{b}\)
1 + ca2 \(\ge2a\sqrt{c}\)
VT \(\ge\)\(2\left(\frac{b\sqrt{a}}{c^3}+\frac{c\sqrt{b}}{a^3}+\frac{a\sqrt{c}}{b^3}\right)\)
\(\ge2\frac{\left(\sqrt[4]{b^2a}+\sqrt[4]{c^2b}+\sqrt[4]{a^2c}\right)^2}{a^3+b^3+c^3}\)
\(\ge2\frac{\left(3\sqrt[12]{a^3b^3c^3}\right)^2}{a^3+b^3+c^3}\)
\(\ge\frac{18}{a^3+b^3+c^3}\)
