1) \(xy\le\frac{\left(x+y\right)^2}{4}\)(cô si) ÁP DỤNG BẤT ĐẲNG THỨC TRÊN với a, b,c>0 TA CÓ
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{\left(a+b\right)^2}{4\left(a+b\right)}+\frac{\left(b+c\right)^2}{4\left(b+c\right)}+\frac{\left(c+a\right)^2}{4\left(c+a\right)}.\)
\(=\frac{a+b}{4}+\frac{b+c}{4}+\frac{c+a}{4}=\frac{2\left(a+b+c\right)}{4}=\frac{a+b+c}{2}.\)
2) Với a,b,c >0 .XÉT \(\frac{a^2}{b}+b\ge2\sqrt{\frac{a^2}{b}.b}=2a\)(bất đẳng thức cô si)
\(\frac{b^2}{c}+c\ge2\sqrt{\frac{b^2}{c}.c}=2b\)
\(\frac{c^2}{a}+a\ge2\sqrt{\frac{c^2}{a}.a}=2c\)
\(\Rightarrow\frac{a^2}{b}+b+\frac{b^2}{c}+c+\frac{c^2}{a}+a\ge2a+2b+2c\)
\(\Leftrightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c\)
(đpcm)
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{ab}{2\sqrt{ab}}+\frac{bc}{2\sqrt{bc}}+\frac{ca}{2\sqrt{ca}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\le\frac{a+b+c}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
Chứng minh bất đẳng thức cô si
Ta có \(\left(x-y\right)^2\ge0\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow x^2+y^2\ge2xy..\)
\(\Leftrightarrow x^2+2xy+y^2\ge2xy+2xy\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Rightarrow\hept{\begin{cases}xy\le\frac{\left(x+y\right)^2}{4}\left(1\right)\\x+y\ge2\sqrt{xy}\left(x,y\ge0\right)\left(2\right)\end{cases}}\)
Ta CM : \(\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Rightarrow\) \(\frac{a+b}{ab}-\frac{4}{a+b}\ge0\)
\(\Rightarrow\)\(\frac{\left(a+b\right).\left(a+b\right)}{ab.\left(a+b\right)}-\frac{4ab}{ab.\left(a+b\right)}\ge0\)
\(\Rightarrow\)\(\frac{\left(a+b\right)^2-4ab}{ab.\left(a+b\right)}\ge0\)
\(\Rightarrow\) \(\frac{a^2+2ab+b^2-4ab}{ab.\left(a+b\right)}\ge0\)
\(\Rightarrow\)\(\frac{a^2-2ab+b^2}{ab.\left(a+b\right)}\ge0\)
\(\Rightarrow\)\(\frac{\left(a-b\right)^2}{ab.\left(a+b\right)}\ge0\) ( LUÔN ĐÚNG )
\(\Rightarrow\)\(\frac{a+b}{ab}\ge\frac{4}{a+b}\)
Do \(\frac{a+b}{ab}\ge\frac{4}{a+b}\)\(\Rightarrow\)\(\frac{ab}{a+b}\le\frac{a+b}{4}\)\(\left(1\right)\)
\(\frac{b+c}{bc}\ge\frac{4}{b+c}\)\(\Rightarrow\)\(\frac{bc}{b+c}\le\frac{b+c}{4}\)\(\left(2\right)\)
\(\frac{a+c}{ac}\ge\frac{4}{a+c}\)\(\Rightarrow\)\(\frac{ac}{a+c}\le\frac{a+c}{4}\)\(\left(3\right)\)
Cộng vế với vế (1),(2) và (3)
\(\Rightarrow\)\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\)\(\le\)\(\frac{a+b}{4}+\frac{b+c}{4}+\frac{a+c}{4}\)
\(\Rightarrow\)\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\)\(\frac{a+b}{4}+\frac{b+c}{4}+\frac{a+c}{4}\)
\(\Rightarrow\)\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\)\(\frac{1}{4}.\left(a+b+b+c+c+a\right)\)
\(\Rightarrow\)\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\)\(\frac{1}{4}.\left(2a+2b+2c\right)\)
\(\Rightarrow\)\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{a+b+c}{2}\)(ĐPCM)
Giải thích : \(\frac{a+b}{ab}\ge\frac{4}{a+b}\Rightarrow\frac{ab}{a+b}\le\frac{a+b}{4}\)
Cũng như : \(\frac{8}{1}>\frac{7}{1}\Rightarrow\frac{1}{8}< \frac{1}{7}\)
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