Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bđt \(\Leftrightarrow\)\(\Sigma_{cyc}\frac{a^2}{2}+\Sigma_{cyc}\frac{a}{bc}\ge\frac{9}{2}\)
mặt khác: \(\Sigma_{cyc}\frac{a}{bc}=\frac{1}{2}\Sigma_{cyc}\left(\frac{b}{ca}+\frac{c}{ab}\right)\ge\Sigma\frac{1}{a}\)\(\Rightarrow\)\(\Sigma_{cyc}\frac{a}{bc}\ge\Sigma_{cyc}\frac{1}{a}\)
do đó cần CM: \(\Sigma_{cyc}\frac{a^2}{2}+\Sigma_{cyc}\frac{1}{a}\ge\frac{9}{2}\) (1)
\(VT_{\left(1\right)}=\Sigma_{cyc}\left(\frac{a^2}{2}+\frac{1}{2a}+\frac{1}{2a}\right)\ge3.\frac{3}{2}=\frac{9}{2}\)
"=" \(\Leftrightarrow\)\(a=b=c=1\)
\(3=ab+bc+ca\ge3\sqrt[3]{abc}\Rightarrow abc\le1\)
\(\Rightarrow VT\le\frac{1}{abc+a^2\left(b+c\right)}+\frac{1}{abc+b^2\left(c+a\right)}+\frac{1}{abc+c^2\left(a+b\right)}\)
\(\Rightarrow VT\le\frac{1}{a\left(ab+bc+ca\right)}+\frac{1}{b\left(ab+bc+ca\right)}+\frac{1}{c\left(ab+bc+ca\right)}\)
\(\Rightarrow VT\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{ab+bc+ca}{3abc}=\frac{1}{abc}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)
\(\Rightarrow\frac{bc}{a^2\left(b+c\right)}+\frac{b+c}{4bc}\ge2\sqrt{\frac{bc}{a^2\left(b+c\right)}\cdot\frac{b+c}{4bc}}=\frac{1}{a}\)
\(\Rightarrow\frac{ca}{b^2\left(c+a\right)}+\frac{c+a}{4ca}\ge2\sqrt{\frac{ca}{b^2\left(c+a\right)}\cdot\frac{c+a}{4ca}}=\frac{1}{b}\)
\(\Rightarrow\frac{ab}{c^2\left(a+b\right)}+\frac{a+b}{4ab}\ge2\sqrt{\frac{ab}{c^2\left(a+b\right)}\cdot\frac{a+b}{4ab}}=\frac{1}{c}\)
Cộng theo vế các bất đẳng thức trên ta được:
\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}+\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Mà\(\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)nên:
\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
hay\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)
Bất đẳng thức xảy ra khi \(a=b=c\)
Áp dụng bđt AM-GM:
\(\frac{1}{a^3\left(b+c\right)}+\frac{a\left(b+c\right)}{4}\ge2\sqrt{\frac{a\left(b+c\right)}{4a^3\left(b+c\right)}}=\frac{1}{a}\)
\(\frac{1}{b^3\left(c+a\right)}+\frac{b\left(c+a\right)}{4}\ge2\sqrt{\frac{b\left(c+a\right)}{4b^3\left(c+a\right)}}=\frac{1}{b}\)
\(\frac{1}{c^3\left(a+b\right)}+\frac{c\left(a+b\right)}{4}\ge2\sqrt{\frac{c\left(a+b\right)}{4c^3\left(a+b\right)}}=\frac{1}{c}\)
Cộng theo vế:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}+\frac{ab+bc+ac}{2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{ab+bc+ac}{2}\)
\(\Leftrightarrow\frac{2}{a^3\left(b+c\right)}+\frac{2}{b^3\left(c+a\right)}+\frac{2}{c^3\left(a+b\right)}\ge ab+bc+ac\) (đpcm)
\("="\Leftrightarrow a=b=c=1\)
giúp với