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\(\frac{ab}{6+a-c}=\frac{ab}{a+b+c+a-c}=\frac{ab}{2a+b}\)
Áp dụng BĐT Cauchy-schwarz ta có:
\(\frac{ab}{2a+b}\le\frac{ab}{9}.\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)=\frac{2b+a}{9}\)
Chứng minh tương tự ta có:
\(\frac{bc}{2b+c}\le\frac{bc}{9}.\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)=\frac{2c+b}{9}\)
\(\frac{ca}{2c+a}\le\frac{ac}{9}.\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)=\frac{2a+c}{9}\)
Dấu " = " xảy ra <=> a=b=c
Cộng vế với vế của 3 BĐT trên ta có:
\(\frac{ab}{6+a-c}+\frac{bc}{6+b-a}+\frac{ac}{6+c-b}\)
\(=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}\le\frac{3\left(a+b+c\right)}{9}=\frac{6}{3}=2\)
Dấu " = " xảy ra <=> a=b=c=2
sr tui ko có câu hỏi tương tự tui chỉ có câu hỏi y hệt thôi Xem câu hỏi
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
\(P=\sum\frac{ab}{a+3b+2c}=\sum\frac{ab}{a+c+b+c+2b}\le\frac{1}{9}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{ab}{2b}\right)=\frac{a+b+c}{6}\)
Dấu "=" có xảy ra tại \(a=b=c\)
mk làm r` đây nhé Câu hỏi của Lê Chí Cường - Toán lớp 9 - Học toán với OnlineMath
\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)
\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)
\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)
\(=\frac{1}{\left(a+b+c\right)a+bc}+\frac{1}{\left(a+b+c\right)b+ac}+\frac{1}{\left(a+b+c\right)c+ab}\)
\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
Vậy VT = VP, đẳng thức được chứng minh
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 :
\(\frac{ac}{c+3a+2b}+\frac{bc}{b+3c+2a}+\frac{ab}{a+3b+2c}\)
\(\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)+\frac{bc}{9}\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{2c}\right)+\frac{ac}{9}\left(\frac{1}{b+c}+\frac{1}{b+a}+\frac{1}{2a}\right)\)
\(=\frac{1}{9}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\frac{1}{9}\left(\frac{ab}{b+c}+\frac{ac}{b+c}\right)+\frac{1}{9}\left(\frac{ac}{a+b}+\frac{bc}{a+b}\right)+\frac{a}{18}+\frac{b}{18}+\frac{c}{18}\)\(=\frac{a+b+c}{6}\)
\(\RightarrowĐPCM\)
Bunhiacopxki:
\(\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)
\(\Rightarrow\dfrac{ab}{a^2+bc+ca}\le\dfrac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)
Tương tự: \(\dfrac{bc}{b^2+ca+ab}\le\dfrac{bc\left(c^2+ca+ab\right)}{\left(ab+bc+ca\right)^2}\)
\(\dfrac{ca}{c^2+ab+bc}\le\dfrac{ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)
\(\Rightarrow VT\le\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\le\dfrac{a^2+c^2+c^2}{ab+bc+ca}\)
\(\Leftrightarrow ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)\le\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\)
Nhân phá và rút gọn 2 vế:
\(\Leftrightarrow a^3b+b^3c+c^3a\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{a^3b+b^3c+c^3a}{abc}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge a+b+c\)
Đúng do: \(\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c\)
+ chứng bất đẳng thức phụ: \(\frac{1}{x+y}\le\frac{1}{4x}+\frac{1}{4y}\left(x,y>0\right)\)
Với \(x,y>0:\left(x-y\right)^2\ge0\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow x^2+2xy+y^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow\frac{x+y}{4xy}\ge\frac{1}{x+y}\Leftrightarrow\frac{1}{x+y}\le\frac{1}{4x}+\frac{1}{4y}\)(đpcm)
Dấu "=" xảy ra \(\Leftrightarrow x-y=0\Leftrightarrow x=y\)
+ Thay \(a+b+c=6\)vào P , ta được: \(P=\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{c+a}\)
Áp dụng bđt chứng minh trên , ta được:\(\frac{1}{a+b}\le\frac{1}{4a}+\frac{1}{4b}\Rightarrow\frac{ab}{a+b}\le ab\left(\frac{1}{4a}+\frac{1}{4b}\right)=\frac{a}{4}+\frac{b}{4}\)
Tương tự như vậy rồi cộng từng vế các bđt , ta được
\(P\le\frac{a}{4}+\frac{b}{4}+\frac{b}{4}+\frac{c}{4}+\frac{c}{4}+\frac{a}{4}=\frac{a+b+c}{2}=\frac{6}{2}=3\)
Dấu "=" xảy ra\(\Leftrightarrow\hept{\begin{cases}a=b=c\\a+b+c=6\end{cases}}\Leftrightarrow a=b=c=2\)
Vậy maxP =3\(\Leftrightarrow a=b=c=2\)
Ta có: \(\frac{ab}{6+a-c}+\frac{bc}{6+b-a}+\frac{ca}{6+c-b}\)
\(=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}\)
Áp dụng BĐT \(\frac{1}{a+b+c}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) với a,b>0
\(VT\le\frac{1}{9}\left(\frac{ab}{a}+\frac{ab}{a}+\frac{ab}{b}\right)+\frac{1}{9}\left(\frac{bc}{b}+\frac{bc}{b}+\frac{bc}{c}\right)+\frac{1}{9}\left(\frac{ca}{c}+\frac{ca}{c}+\frac{ca}{a}\right)=\frac{1}{3}\left(a+b+c\right)=2\)