Ta có:
\(\frac{a^2}{b^2}+1\ge2.\frac{a}{b}\)
\(\frac{b^2}{c^2}+1\ge2.\frac{b}{c}\)
\(\frac{c^2}{a^2}+1\ge2.\frac{c}{a}\)
Cộng vế theo vế ta được
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+3\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-3\)
\(\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\sqrt{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}-3=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)
Dấu = xảy ra khi a = b = c
Ta co: \(\frac{a^2}{b^2}\ge\frac{a}{b}\); \(\frac{b^2}{c^2}\ge\frac{b}{c}\);\(\frac{c^2}{a^2}\ge\frac{c}{a}\)\(\Rightarrow dpcm\)
ta có bất đẳng thức: \(x^2\)\(+\)\(y^2\)\(>=2xy\)
chứng minh: \(x^2\)\(+\)\(y^2\)\(-\)\(2xy\)\(>=0\)
\(=>\)(\(x\)\(-y\))^\(2\)\(>=0\)(luôn đúng với mọi a,b)
vậy \(x^2\)\(+\)\(y^2\)\(>=2xy\)
áp dụng bát đăng thức trên ta có:
\(\frac{a^2}{b^2}\)\(+\)\(\frac{b^2}{c^2}\)\(>=2.\)\(\frac{a}{b}\)\(.\)\(\frac{b}{c}\)\(=\)\(2.\)\(\frac{a}{c}\)
\(\frac{a^2}{b^2}\)\(+\)\(\frac{c^2}{a^2}\)\(>=\)\(2.\)\(\frac{a}{b}\)\(.\)\(\frac{c}{a}\)\(=\)\(2.\)\(\frac{c}{b}\)
\(\frac{b^2}{c^2}\)\(+\)\(\frac{c^2}{a^2}\)\(>=\)\(2.\)\(\frac{b}{c}\)\(.\)\(\frac{c}{a}\)\(=\)\(2.\)\(\frac{b}{a}\)
cộng từng vế ba bất đẳng thức trên ta được:
\(\frac{a^2}{b^2}\)\(+\)\(\frac{b^2}{c^2}\)\(+\)\(\frac{c^2}{a^2}\)\(+\)\(\frac{a^2}{b^2}\)\(+\)\(\frac{b^2}{c^2}\)\(+\)\(\frac{c^2}{a^2}\)\(>=\)\(2\)(\(\frac{a}{c}\)+\(\frac{b}{a}\)+\(\frac{c}{b}\))
\(2\)(\(\frac{a^2}{b^2}\)+\(\frac{b^2}{c^2}\)\(+\)\(\frac{c^2}{a^2}\))\(>=\)\(2\).(\(\frac{a}{c}\)\(+\)\(\frac{b}{a}\)\(+\)\(\frac{c}{b}\))
\(=>\)\(\frac{a^2}{b^2}\)\(+\)\(\frac{b^2}{c^2}\)\(+\)\(\frac{c^2}{a^2}\)\(>=\)\(\frac{a}{c}\)\(+\)\(\frac{b}{a}\)\(+\)\(\frac{c}{b}\)(đpcm)
k cho mình nhé