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\(P=\frac{ab}{6-c}+\frac{bc}{6-a}+\frac{ac}{6-b}\)
\(P=\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\)
Ta có: \(\hept{\begin{cases}ab\le\frac{\left(a+b\right)^2}{4}\\bc\le\frac{\left(b+c\right)^2}{4}\\ac\le\frac{\left(a+c\right)^2}{4}\end{cases}}\)(bđt AM-GM)
\(\Rightarrow P\le\frac{\left(a+b\right)^2}{4\left(a+b\right)}+\frac{\left(b+c\right)^2}{4\left(b+c\right)}+\frac{\left(a+c\right)^2}{4\left(a+c\right)}=\frac{a+b+b+c+a+c}{4}=3\)
\("="\Leftrightarrow a=b=c=2\)
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\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+3\ge7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le3\)Áp dụng BĐT AM-GM ta có :
\(A=\frac{1}{\sqrt{a^3+b^3+1}}+\frac{1}{\sqrt{b^3c^3+1+1}}+\frac{4\sqrt{3}}{c^6+1+2a^3+8}\)
\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{4\sqrt{3}}{2c^3+2a^3+8}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+4}\)
\(=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+1+1+1+1}\)
\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{6\sqrt{ac}}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{1}{\sqrt{3ac}}\)\(=\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)
\(\le\frac{1}{\sqrt{3}}\sqrt{3\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}=\sqrt{\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}\le\sqrt{3}\) (Bunhiacopxki)
Dấu "=" xảy ra\(\Leftrightarrow a=b=c=1\)
PS : Thánh cx đc phết ha; chế đc bài này tui mới khâm phục :)))
nó ko chém đâu anh nó chép trong toán tuổi thơ đấy,thk này khốn nạn lắm
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\(P=\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\)
\(P\le\dfrac{ab}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{bc}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{ca}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
\(\Leftrightarrow P\le\dfrac{1}{2}\left(a+b+c\right)=3\)
\(P_{max}=3\) khi \(a=b=c\)
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
mk làm r` đây nhé Câu hỏi của Lê Chí Cường - Toán lớp 9 - Học toán với OnlineMath
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(P=\frac{1}{\left(a+2\right)+\left(a+2\right)+\left(b+2\right)}+\frac{1}{\left(b+2\right)+\left(b+2\right)+\left(c+2\right)}+\frac{1}{\left(c+2\right)+\left(c+2\right)+\left(a+2\right)}\)
\(\le\frac{1}{9}\left(\frac{2}{a+2}+\frac{1}{b+2}\right)+\frac{1}{9}\left(\frac{2}{b+2}+\frac{1}{c+2}\right)+\frac{1}{9}\left(\frac{2}{c+2}+\frac{1}{a+2}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)\)
Dễ dàng cm BĐT \(\frac{1}{x+1}+\frac{1}{y+1}\ge\frac{2}{1+\sqrt{xy}}\)
\(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=\frac{1}{2}\left(\frac{1}{1+\frac{a}{2}}+\frac{1}{1+\frac{b}{2}}+\frac{1}{1+\frac{c}{2}}\right)\)
\(\le\frac{1}{2}.\frac{3}{1+\sqrt[3]{\frac{abc}{8}}}=\frac{3}{4}\Rightarrow P\le\frac{1}{4}\)
Xảy ra khi \(a=b=c=2\)
À viết ngược dấu BĐT phụ r` :v
\(\frac{1}{1+x}+\frac{1}{1+y}\le\frac{2}{1+\sqrt{xy}}\) mới đúng nhé :v
\(\Leftrightarrow\frac{\left(\sqrt{xy}-1\right)\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x+1\right)\left(y+1\right)\left(1+\sqrt{xy}\right)}\le0\)
+ 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\(\Leftrightarrow a=b=c=2\)