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Ta có: \(\frac{2a^3}{a^6+bc}\le\frac{2a^3}{2a^3\sqrt{bc}}=\frac{1}{\sqrt{bc}}\\ \)
CMTT: \(\frac{2b^3}{b^6+ca}\le\frac{1}{\sqrt{ca}}\)
\(\frac{2c^3}{c^6+ab}\le\frac{1}{\sqrt{ab}}\)
\(\Rightarrow\frac{2a^3}{a^6+bc}+\frac{2b^3}{b^6+ca}+\frac{2c^3}{c^6+ab}\le\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}+\frac{1}{\sqrt{ab}}\)\(=\) \(\frac{\sqrt{bc}}{bc}+\frac{\sqrt{ac}}{ac}+\frac{\sqrt{ab}}{ab}\)
\(\le\frac{a+c}{2ac}+\frac{b+c}{2bc}+\frac{a+b}{2ab}=\frac{2\left(ab+bc+ca\right)}{2abc}=\frac{ab+bc+ca}{abc}\) \(\le\frac{a^2+b^2+c^2}{abc}=\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\left(đpcm\right)\)
Dấu bằng xảy ra khi : a = b = c =1
#)Giải :
Ta có :
\(\hept{\begin{cases}\frac{ab}{b+c+a+b}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\\frac{bc}{a+b+a+c}\le\frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\\\frac{ac}{b+c+a+b}\le\frac{ac}{4}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)\end{cases}}\)
\(\Rightarrow VT\le\frac{1}{a+b}.\left(\frac{bc}{4}+\frac{ac}{4}\right)+\frac{1}{a+c}.\left(\frac{bc}{4}+\frac{ab}{4}\right)+\frac{1}{b+c}.\left(\frac{ac}{4}+\frac{ab}{4}\right)\)
\(=\frac{1}{a+b}.\frac{c\left(a+b\right)}{4}+\frac{1}{a+c}.\frac{b\left(a+c\right)}{4}+\frac{1}{b+c}.\frac{a\left(b+c\right)}{4}\)
\(=\frac{c}{4}+\frac{b}{4}+\frac{a}{4}\)
\(\Rightarrow\frac{a+b+c}{4}\)
\(\Rightarrowđpcm\)
Áp đụng bất đẳng thức Cauchy-Schwartz , ta có :
\(\frac{ab}{a+3b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)\)
Tương tự , ta có:
\(\frac{bc}{b+3c+2a}=\frac{bc}{\left(a+b\right)+\left(a+c\right)+2c}\le\frac{bc}{9}\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{2c}\right)\)
\(\frac{ac}{c+3a+2b}=\frac{ac}{\left(b+c\right)+\left(b+a\right)+2b}\le\frac{ac}{9}\left(\frac{1}{b+c}+\frac{1}{b+a}+\frac{1}{2a}\right)\)
Cộng vế theo vế ta có :
\(\frac{ac}{c+3a+2b}+\frac{bc}{b+3c+2a}+\frac{ab}{a+3b+2c}\)
\(\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)+\frac{bc}{9}\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{2c}\right)+\frac{ac}{9}\left(\frac{1}{b+c}+\frac{1}{b+a}+\frac{1}{2a}\right)\)
\(=\frac{1}{9}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\frac{1}{9}\left(\frac{ab}{b+c}+\frac{ac}{b+c}\right)+\frac{1}{9}\left(\frac{ac}{a+b}+\frac{bc}{a+b}\right)+\frac{a}{18}+\frac{b}{18}+\frac{c}{18}\)\(=\frac{a+b+c}{6}\)
\(\RightarrowĐPCM\)
\(P=\sum\frac{ab}{a+3b+2c}=\sum\frac{ab}{a+c+b+c+2b}\le\frac{1}{9}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{ab}{2b}\right)=\frac{a+b+c}{6}\)
Dấu "=" có xảy ra tại \(a=b=c\)
\(\frac{a^3}{bc}+\frac{b^3}{ca}=\frac{a^4}{abc}+\frac{b^4}{abc}\ge\frac{\left(a^2+b^2\right)^2}{2abc}\ge\frac{2ab\left(a^2+b^2\right)}{2abc}=\frac{a^2+b^2}{c}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b\)
viết các bđt tương tự rồi cộng vế theo vế là được
Ối,không ngờ đề gắt ~v
Theo Cô si,ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{3}{\sqrt[3]{xyz}}\ge\frac{3}{\frac{x+y+z}{3}}=\frac{9}{x+y+z}\)
Suy ra \(\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Áp dụng vào,ta có: \(\frac{1}{a+2b+3c}=\frac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\)
\(\le\frac{1}{9}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{b+c}\right)\)
Chứng minh tương tự và cộng theo vế:
\(VT\le\frac{1}{9}\left[\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)+2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\right]\)
\(=\frac{1}{9}\left[3\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\right]=\frac{1}{3}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
Lại có BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Áp dụng vào,ta có: \(VT\le\frac{1}{3}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(\le\frac{1}{12}\left[2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Nhân abc vào mỗi vế : \(VT.abc\le\frac{1}{6}\left(ab+bc+ca\right)=\frac{abc}{6}\)
Chia cả hai vế cho abc (vì a,b,c dương nên abc khác 0): \(VT\le\frac{1}{6}< \frac{3}{16}\)(đpcm)
Cũng không biết đúng hay sai nữa :v
Ta có : \(p=\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(a+c\right)}+\frac{ab}{c^2\left(a+b\right)}\)
Áp dụng bất đẳng thức AM - GM ta có :
\(\frac{bc}{a^2\left(b+c\right)}+\frac{b+c}{4bc}\ge2\sqrt{\frac{bc}{a^2\left(b+c\right)}.\frac{b+c}{4ab}}=\frac{1}{a}\)
\(\frac{ac}{b^2\left(a+c\right)}+\frac{a+c}{4ac}\ge4\sqrt{\frac{ac}{b^2\left(a+c\right)}.\frac{a+c}{4ac}}=\frac{1}{b}\)
\(\frac{ab}{c^2\left(a+b\right)}+\frac{a+b}{4ab}\ge2\sqrt{\frac{ab}{c^2\left(a+b\right)}.\frac{a+b}{4ab}}=\frac{1}{c}\)
Cộng vế với vế ta được \(p+\frac{1}{4c}+\frac{1}{4a}+\frac{1}{4b}+\frac{1}{4a}+\frac{1}{4c}+\frac{1}{4b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow p+\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Rightarrow p\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\ge3\sqrt[3]{\frac{1}{2a.2b.2c}}=\frac{3}{\sqrt[3]{8abc}}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Xét: \(\frac{bc}{a^2b+ca^2}=\frac{bc}{a\cdot abc\cdot\frac{1}{c}+a\cdot abc\cdot\frac{1}{b}}=\frac{b^2c^2}{ab+ca}\)(*)
Tương tự với (*) ta có: \(\hept{\begin{cases}\frac{ca}{b^2c+ab^2}=\frac{c^2a^2}{ab+bc}\\\frac{ab}{c^2a+bc^2}=\frac{a^2b^2}{ca+bc}\end{cases}}\)
\(\Rightarrow\Sigma_{cyc}\frac{bc}{a^2b+ca^2}=\Sigma_{cyc}\frac{b^2c^2}{ab+ca}\)
Ta thấy\(\Sigma_{cyc}\frac{b^2c^2}{ab+ca}\) có dạng: \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2}\left(a+b+c\right)\)
Bước cuối Cô-si ba số và kết hợp điều kiện abc=1 là xong
đề bạn sai dấu rồi nha
uk đúng rồi mk sorry vậy nếu là dấu nhỏ hơn hoặc bằng bạn có thể giải giúp mk ko